We're sorry, this computer has been flagged for suspicious activity.

If you are a member, we ask that you confirm your identity by entering in your email.

You will then be sent a link via email to verify your account.

If you are not a member or are having any other problems, please contact customer support.

Thank you for your cooperation

Forgot password? New user? Sign up

Existing user? Log in

Integration Tricks

Already have an account? Log in here.

Many challenging integration problems can be solved surprisingly quickly by simply knowing the right technique to apply. While finding the right technique can be a matter of ingenuity, there are a dozen or so techniques that permit a more comprehensive approach to solving definite integrals.

Manipulations of definite integrals may rely upon specific limits for the integral, like with odd and even functions , or they may require directly changing the integrand itself, through some type of substitution . However, most integrals require a combination of techniques, and many of the more complicated approaches, like interpretation as a double integral , require multiple steps to reduce the expression.

Consider, for instance, the antiderivative

\[\displaystyle\int e^{- x^2} \, dx.\]

This is known as the Gaussian integral, after its usage in the Gaussian distribution , and it is well known to have no closed form. However, the improper integral

\[I = \int_0^\infty e^{- x^2} \, dx\]

may be evaluated precisely, using an integration trick . In fact, its value is given by the polar integral

\[I^2 = \int_0^\infty \int_0^\infty e^{-x^2} e^{-y^2} \, dy\, dx = \int_0^{\pi/2} \int_0^\infty r e^{-r^2} \, dr\, d\theta.\]

Without such a method for exact evaluation of the integral, the Gaussian (normal) distribution would be significantly more complicated. Such integrals appear throughout physics , statistics , and mathematics .

Odd and Even Functions

Reflections, cyclic points, inverse functions, integration by parts, trigonometric substitutions, weierstrass substitution, taylor series, differentiation under the integral sign, changing to a double integral, harmonic functions.

An odd function \(o(x)\) satisfies \(o(x) = - o(-x)\) for all \(x\). Therefore, for any finite \(t\),

\[\int_{-t}^t o(x) \, dx = 0.\]

An even function \(e(x)\) satisfies \(e(x) = e(-x)\) for all \(x\). Therefore, for any \(t\),

\[\int_{-t}^t e(x) \, dx = 2 \int_0^t e(x) \, dx.\]

Evaluate \[\int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx.\] Notice that the integrand is an odd function. So, \[\begin{align*} \int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx &= \int_{-1}^{1} \frac{(-x)^3-2(-x)}{\sqrt{(-x)^4+1}} \, d(-x) \\ &= - \int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx \\ &= 0. \end{align*}\] The final equivalence comes from the fact that the integral is equal to the negative of itself. Therefore, it is \(0\). \(_\square\)

\[ \large \int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ x }^{ 2 }+{ x }^{ 4 } } \, dx } = \, ? \]

A similar method to the above is to reverse the interval of integration: to "integrate backwards." For a function \(f\) and real numbers \(a < b\),

\[\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.\]

Instead of the function being centered at \(0\), the function is now centered at \(\tfrac{a+b}{2}\). Then,

\[\int_a^b f(x) \, dx = \frac{1}{2} \int_a^b f(x) + f(a+b-x) \, dx.\]

Evaluate \[ \int_3^7 \frac{\ln(x+2)}{\ln\big(24+10x-x^2\big)} \, dx.\] We have \[\begin{align*} \int_3^7 \frac{\ln(x+2)}{\ln\big(24+10x-x^2\big)} \, dx &= \int_3^7 \frac{\ln(x+2)}{\ln(x+2) + \ln(12-x)} \, dx \\\\ &= \frac{1}{2} \int_3^7 \frac{\ln(12-x) + \ln(x+2)}{\ln(x+2) + \ln(12-x)} \, dx \\\\ &= 2.\ _\square \end{align*}\]

Let \(f(x)\) be a real-valued function continuous on \(\left[0,2\right]\) such that \(f(x)=f(2x)\) for all \(x\). If

\[\int_0^1 f(x) dx = 100,\]

then what is the value of

\[\int_0^2 f(x)dx ?\]

Suppose the function \(f\) has bounded antiderivative on \([0, \, \infty]\). Then, via the u-substitution \(x \mapsto \tfrac{1}{x}\),

\[\int_0^\infty f(x) \, dx = \frac{1}{2} \int_0^\infty f(x) + \frac{f\big(\frac{1}{x}\big)}{x^2} \, dx.\]

Evaluate \[ \int_0^\infty \frac{\ln (2x)}{1 + x^2} \, dx.\] We have \[\begin{align} \int_0^\infty \frac{\ln (2x)}{1 + x^2} \, dx &= \frac{1}{2} \int_0^\infty \frac{\ln(2x) + \ln\big(2x^{-1}\big)}{1 + x^2} \, dx \\\\ &= \frac{2\ln 2}{2} \int_0^\infty \frac{1}{1 + x^2} \, dx \\\\ &= \frac{\pi \ln 2}{2}.\ _\square \end{align}\]

Evaluate the integral \(\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^2+2x+4} \, dx.\)

Round your answer to three decimal places.

This section is currently incomplete. Let's join hands to build this wiki. Feel free to add anything you know about this topic!

There are more transformations than simply reflections and inversions that maintain the interval of integration, but they are not as common.

Suppose the function \(f\) is one-to-one and increasing. Then, a geometric equivalence may be established:

\[\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(x) \, dx = bf(b) - af(a).\]

Suppose the function \(f\) is one-to-one and decreasing. Then, another geometric equivalence may be established:

\[\int_a^b f(x) \, dx - \int_{f(b)}^{f(a)} f^{-1}(x) \, dx = (b-a)f(b) - a\big(f(a)-f(b)\big).\]

Let \(f(x)\) be a one-to-one continuous function such that \(f(1)=4\) and \(f(6)=2\), and assume \(\displaystyle \int_1^6 f(x) \, dx = 15\). Calculate \(\displaystyle \int_2^4 f^{-1}(x) \, dx\). The region bounded by \(f\), \(x = 1\), and \(y = 2\) must have area \(5\), implying the integral in question corresponds to the area \(5 + 1 \cdot (4 - 2) = \boxed{7}\). The above formula for decreasing functions provides the same answer. \(_\square\)

Integration by parts provides a way to change the integrand directly, and like the exploration of inverse functions, it is a geometric statement. However, this is a statement about the geometry of calculus operators, and any visualization of it would lie in an entirely different space. However, the same intuition can apply. Integration by parts is a very powerful tool, and many problems on this page could be solved by this (and more elementary methods) without the need for anything more complicated.

Integration by parts states that for any differentiable functions \(u(x)\) and \(v(x)\), the following equivalence holds:

\[ \int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) u'(x) \, dx. \]

This can be thought of as a "backwards" application of the product rule .

Evaluate \[\int_1^7 \ln(1 + x) \, dx.\] Let \(u(x) = \ln(1 + x)\) and \(v(x) = x + 1\). Then, \[\begin{align} \int_1^7 \ln(1 + x) \, dx &= \big[(x+1)\ln(x+1)\big]_1^7 - \int_1^7 \frac{x+1}{x+1} \, dx \\ &= 8\ln 8 - 2\ln 2 - 6 \\ &= 22 \ln 2 - 6.\ _\square \end{align}\]
The (adjusted) beta function \(B(m, \, n)\) is defined for nonnegative integers \(m\) and \(n\) as \[B(m, \, n) = \int_0^1 x^{m}(1-x)^{n} \, dx.\] Find a closed form for \(B(m, \, n)\). Note that \(B(0, \, n) = \frac{1}{n+1}\). Now, suppose \(n\) is fixed, and note for \(m > 0\), \[\begin{align} B(m, \, n) &= \int_0^1 x^m (1-x)^n \, dx \\ &= 0 - \frac{m}{n+1} \int_0^1 x^{m-1} \cdot \left(-(1-x)^{n+1}\right) \, dx \\ &= \frac{m}{n+1} B(m-1, \, n+1). \end{align}\] Thus, \(B(m, \, n) = \frac{m}{n+1} \cdot \frac{m-1}{n+2} \cdots \frac{1}{n+m} B(0, n+m) = \frac{m! n!}{(m+n+1)!}.\) \(_\square\) Learn more about the beta function (with correctly off-set indices) here .

\[ \int_0^1 \left(1-x^2\right)^9 x^9 \, dx \]

Let \(I\) denote the value of the integral above. What is the sum of digits of \(I^{-1}?\)

When solving integrals with trigonometric functions, trigonometric identities create shortcuts! The Integration Of Trigonometric Functions wiki goes into this in detail, but below are a few examples.

The first identity is \(\sin^2x+\cos^2x=1.\)

We have \[\begin{align} \int(\sin x+\cos x)^2\, dx &=\int\left(\sin^2x+2\sin x\cos x+\cos^2x\right)\, dx\\ &=\int(1+\sin 2x)\, dx\\ &=x-\dfrac{1}{2}\cos 2x+C, \end{align}\] where \(C\) is the constant of integration.

Here is an example of a less obvious application of that identity:

We have \[\begin{align} \int_0^\frac{\pi}{2}\sin^5x\, dx &=\int_0^\frac{\pi}{2}\sin x\left(1-\cos^2x\right)^2\, dx\\ &=-\int_1^0\left(1-u^2\right)^2\, du\\ &=\int_0^1\left(1-u^2\right)^2\, du\\ &=\frac{8}{15}. \end{align}\]

Other examples that can be used are the double-angle formulas , which can be used in the integrals of \(\sin^2\theta\) and \(\cos^2\theta,\) as well as others.

We have \[\begin{align} \int\frac{\cos 2x}{\sin x+\cos x}\, dx &=\int\frac{\cos^2x-\sin^2x}{\sin x+\cos x}\, dx\\ &=\int(\cos x-\sin x)\, dx\\ &=\sin x+\cos x+C, \end{align}\] where \(C\) is the constant of integration.

Finally, the product-to-sum identities help to solve complex integrals.

We have \[\begin{align} \int\sin 2015x \sin 2016x \, dx &=\dfrac{1}{2}\int(\cos x-\cos 4031x)\, dx\\ &=\dfrac{\sin x}{2}-\dfrac{\sin 4031x}{8062}+C, \end{align}\] where \(C\) is the constant of integration.

One of the most powerful substitutions using trigonometric functions is the Weierstrass substitution of \(t=\tan\frac{\theta}{2}.\) This is most easily seen in rational functions involving trigonometric functions. Through trigonometric identities and manipulation \(\sin\theta=\frac{2t}{1+t^2},\) \(\cos\theta=\frac{1-t^2}{1+t^2},\) and \(d\theta=\frac{2\ dt}{1+t^2}.\) This can often transform the integral into the integral of a rational function, as seen in the following example:

Find the value of \(\displaystyle\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}.\) Use the Weierstrass substitution \(\cos x=\frac{1-t^2}{1+t^2}\) and \(dx=\frac{2dt}{1+t^2}:\) \[\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\ =\int_0^1\frac{\frac{2dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}=\int_0^1\frac{2}{t^2+3}dt.\] This integral can be finished by a \(u\)-substitution and application of the derivative of arctangent. \(_\square\)

The Weierstrass substitution can also be applied to a rather common integral, \(\displaystyle\int\csc x\ dx.\) A common method of solving this question is a "clever" multiplication by \(\frac{1}{1},\) but the Weierstrass substitution is easier to apply.

We have \[\int\csc \theta\, d\theta=\int\frac{d\theta}{\sin\theta}=\int\frac{\hspace{3mm} \frac{2dt}{1+t^2}\hspace{3mm} }{\frac{2t}{1+t^2}}=\int\frac{dt}{t}=\ln t+C,\] where \(C\) is the constant of integration. All that's needed is to re-substitute \(t=\tan\frac{\theta}{2}\) to obtain the final value of \[\ln\left(\tan\frac{\theta}{2}\right)+C=-\ln(\cot x+\csc x)+C=\ln\frac{\sin\theta}{\cos\theta+1}+C.\]

A final use of the Weierstrass substitution is the "reverse Weierstrass substitution," which involves simplifying the integral of a rational function with trigonometry.

Find the value of \(\displaystyle\int_0^1\dfrac{\arcsin\frac{2x}{1+x^2}}{1+x^2}\, dx\). Recall that in the Weierstrass substitution, \(\frac{2x}{1+x^2}=\sin\theta\) and \(\frac{2dx}{1+x^2}=d\theta.\) Then the above can be transformed to \[\displaystyle\int_0^\frac{\pi}{2}\frac{\sin^{-1}(\sin\theta)}{2}\, d\theta=\displaystyle\int_0^\frac{\pi}{2}\frac{\theta}{2}\, d\theta.\] This is the integral of a polynomial, which evaluates to \(\frac{\pi^2}{16}.\) \(_\square\)

\[\int_0^1\frac{x^4\left(1-x^2\right)^5}{\left(1+x^2\right)^{10}}\, dx=A\]

Given the above, find \(\frac{1}{A}.\)

Main article: Taylor Series

Some functions like \(\tfrac{1}{1-x}\), \(\ln(1 - x)\), \(\arctan x\), and \(e^x\) have nice Taylor expansions that, together with term-by-term integration, can lead to a closed-form answer. The monotone convergence theorem states that in most cases where the integral does exist (as should generally be the case when evaluating an integral), the summation and integral may be interchanged. For more information, see Double Integrals .

Evaluate \[ \int_0^1 \ln x \ln(1 - x) \, dx.\] Note \[ \int_0^1 \ln x \ln(1-x) \, dx = - \int_0^1 \sum_{k = 1}^\infty \frac{x^k \ln x}{k} \, dx.\] Since the monotone convergence theorem applies here, this is equal to \[\begin{align} - \sum_{k = 1}^\infty \int_0^1 \frac{x^k \ln x}{k} \, dx &= \sum_{k = 1}^\infty \frac{1}{k (k+1)^2} \\ &= \sum_{k = 1}^\infty \frac{1}{k(k + 1)} - \sum_{k = 1}^\infty \frac{1}{(k + 1)^2} \\ &= 2 - \frac{\pi^2}{6}.\ _\square \end{align}\]
Main article: Differentiation under the integral sign

Differentiating under the integral sign is a useful method for evaluating certain integrals which might be harder using other methods. This method of integrating was so frequently used by Richard Feynman that it is often referred to as Feynman's integration trick .

\[\dfrac{d}{dx} \displaystyle \int_{g(x)}^{h(x)} f(x,t)dt = f\big(x,h(x)\big)\dfrac{d}{dx}h(x) - f\big(x,g(x)\big)\dfrac{d}{dx}g(x) + \displaystyle \int_{g(x)}^{h(x)}\dfrac{\partial}{\partial x}f(x,t)dt. \]
Compute \[ \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-7x}}{x} \, dx. \] Let \( I(a) = \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-ax}}{x}\, dx.\) Upon differentiating under the integral sign, the equation becomes \[\begin{align} \dfrac{\partial I}{\partial a} &= \displaystyle \int_{0}^{\infty} \dfrac{\partial }{\partial a}\dfrac{e^{-5x} - e^{-ax}}{x}\, dx \\ &= \displaystyle \int_{0}^{\infty} \dfrac{0 -(-xe^{-ax})}{x}\, dx \\ &= \displaystyle \int_{0}^{\infty} e^{-ax}\, dx. \end{align}\] Now, integrating with respect to \(x\) yields the following: \[ \dfrac{\partial I}{\partial a} = \left[\dfrac{e^{-ax}}{a}\right]_{\infty}^{0} \implies \dfrac{\partial I}{\partial a} = \dfrac{1}{a} .\] Integrating both sides with respect to \(a,\) \[ I(a) = \displaystyle \int \dfrac{1}{a}\, da + C \Rightarrow I(a) = \ln a + C, \qquad (1) \] where \(C\) is the constant of integration. Notice that \( I(5) = \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-5x}}{x}\, dx = \int_{0}^{\infty} 0\, dx = 0. \) Substituting these values in \((1)\) gives \[ 0 = \ln 5 + C \implies C = -\ln 5 \implies I(a) = \ln\dfrac{a}{5}.\] To obtain the required integral, substitute \(a = 7\): \[ I(7) = \ln\dfrac{7}{5} \implies \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-7x}}{x}\, dx = \ln\dfrac{7}{5}. \ _\square\]
Compute \[ \displaystyle \int_{0}^{\infty} \dfrac{\sin x}{x}\, dx. \] Let \( I(a) = \displaystyle \int_{0}^{\infty} e^{-ax}\dfrac{\sin x}{x}\, dx. \) Then differentiating under the integral sign gives \[ \dfrac{\partial I(a)}{\partial a} = \displaystyle \int_{0}^{\infty} \dfrac{\partial}{\partial a} e^{-ax}\dfrac{\sin x}{x}\, dx= -\displaystyle \int_{0}^{\infty} e^{-ax}\sin x\, dx. \] Integrating with respect to \(x,\) \[ \dfrac{\partial I(a)}{\partial a} = \left[\dfrac{e^{-ax}\left(a\sin x + \cos x\right)}{a^{2}+1}\right]_{0}^{\infty} = -\dfrac{1}{1+a^{2}}. \] Integrating with respect to \(a,\) \[ I(a) = \displaystyle \int -\dfrac{1}{1+a^{2}}da = -\tan^{-1} a + C, \] where \(C\) is the constant of integration. Now, \[ \displaystyle \lim_{a\rightarrow \infty} = \displaystyle \int_{0}^{\infty} \displaystyle \lim_{a\rightarrow \infty} e^{-ax}\dfrac{\sin x}{x}dx = 0 .\] Using the above information, \[ 0 = -\displaystyle \lim_{a \rightarrow \infty} \tan^{-1} a + C \Rightarrow C = \dfrac{\pi}{2} \Rightarrow I(a) = \dfrac{\pi}{2} - \tan^{-1} a. \] To get our integral, we let \(a = 0\) to obtain \[ I(0) = \dfrac{\pi}{2} - \tan^{-1} 0 = \dfrac{\pi}{2}. \] Therefore, \( \displaystyle \int_{0}^{\infty} \dfrac{\sin x}{x}\, dx = \dfrac{\pi}{2} . \ _\square\)

\[ \int_0^1 x^{50} (\ln x)^{150} \, dx \]

If the value of the integral above is equal to

\[ \dfrac{A!}{B^C}, \]

where \(A,B,\) and \(C\) are positive integers, find the value of \(A+B+C\).

Bonus : Generalize \( \displaystyle \int_0^1 x^{m} (\ln x)^{n} \, dx \).

Find the value of the following integral:

\[ \displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1} \pi x - \tan^{-1}x\right)\,dx.\]

Sometimes, the integrand looks like it has already been integrated. This may signal that the integral is better interpreted as a double integral . There are more possibilities for \(u\)-substitutions when two variables can be manipulated (polar, skewed, etc), and simply changing the order of integration may suffice to simplify the integral.

In many ways, this is a dual method to differentiation under the integral sign. The main difference is that the extra variable is interpreted inside of the variable rather than outside of it. In most cases where one works, both could work; in some cases, only one approach works nicely, so it is good to know both.

Suppose \(a\) and \(b\) are real numbers, \(f\) a function, and \(I = \int_a^b f(x) \, dx.\) If \(f(x) = g(x, s) - g(x, r)\) for some constants \(r\) and \( s\), then \[I = \int_a^b \int_r^s \frac{\partial}{\partial t} g(x,t) \, dt \, dx.\]

Supposing Fubini's theorem holds, the order of integration may be swapped or otherwise altered.

Evaluate \[\int_0^{\infty} \frac{e^{\pi x} - e^{x}}{x(e^{\pi x}+1)(e^{x}+1)} \, dx.\] We have \[ \begin{align*} \int_0^{\infty} \frac{e^{\pi x} - e^{x}}{x(e^{\pi x}+1)(e^{x}+1)} \, dx &= \int_0^\infty \frac{1}{x (e^x + 1)} - \frac{1}{x (e^{\pi x} + 1)} \, dx \\ &= \int_0^\infty \int_1^\pi \frac{e^{tx}}{(e^{tx} + 1)^2} \, dt \, dx \\ &= \int_1^\pi \int_0^\infty \frac{e^{tx}}{(e^{tx} + 1)^2} \, dx \, dt \\ &= \int_1^\pi \left[ - \frac{1}{t (e^{tx} + 1)} \right]_0^\infty \, dt \\ &= \int_1^\pi \frac{1}{(1 + 1)t} \, dt \\ &= \frac{\ln \pi}{2}.\ _\square \end{align*} \]

In complex analysis, a harmonic function is a real-valued function that is the real or imaginary part of a complex-differentiable function. In multivariable calculus, it is a function \(f(x, \, y)\) such that \(\left(\frac{\partial}{\partial x}\right)^2 f + \left( \frac{\partial}{\partial y}\right)^2f = 0.\) Generally, such facts from fields afar are not applicable to the evaluation of real integrals; however, the harmonic functions have a special property that greatly simplifies integration over circles .

Suppose \(f\) is a bivariate harmonic function, \((a, \, b)\) is a point in the plane, and \(r\) is a positive real number. Then, \[ \int_0^{2\pi} f(a + r\cos\theta, \, b + r\sin\theta) \, d\theta = 2\pi f(a, \, b).\]
Evaluate \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx.\] Consider the function \(f(x, \, y) = e^x \cos y.\) Note that \[ \left(\frac{\partial}{\partial x}\right)^2 f + \left(\frac{\partial}{\partial y}\right)^2 f = e^x \cos y + e^x (- \cos y) = 0.\] Therefore, \(f\) is a harmonic function, and it follows that \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx = 2\pi e^0 \cos(0) = 2\pi.\ _\square\]

Problem Loading...

Note Loading...

Set Loading...

Section 7.7 : Integration Strategy

We’ve now seen a fair number of different integration techniques and so we should probably pause at this point and talk a little bit about a strategy to use for determining the correct technique to use when faced with an integral.

There are a couple of points that need to be made about this strategy. First, it isn’t a hard and fast set of rules for determining the method that should be used. It is really nothing more than a general set of guidelines that will help us to identify techniques that may work. Some integrals can be done in more than one way and so depending on the path you take through the strategy you may end up with a different technique than somebody else who also went through this strategy.

Second, while the strategy is presented as a way to identify the technique that could be used on an integral also keep in mind that, for many integrals, it can also automatically exclude certain techniques as well. When going through the strategy keep two lists in mind. The first list is integration techniques that simply won’t work and the second list is techniques that look like they might work. After going through the strategy and the second list has only one entry then that is the technique to use. If, on the other hand, there is more than one possible technique to use we will then have to decide on which is liable to be the best for us to use. Unfortunately, there is no way to teach which technique is the best as that usually depends upon the person and which technique they find to be the easiest.

Third, don’t forget that many integrals can be evaluated in multiple ways and so more than one technique may be used on it. This has already been mentioned in each of the previous points but is important enough to warrant a separate mention. Sometimes one technique will be significantly easier than the others and so don’t just stop at the first technique that appears to work. Always identify all possible techniques and then go back and determine which you feel will be the easiest for you to use.

Next, it’s entirely possible that you will need to use more than one method to completely do an integral. For instance, a substitution may lead to using integration by parts or partial fractions integral.

Finally, in my class I will accept any valid integration technique as a solution. As already noted there is often more than one way to do an integral and just because I find one technique to be the easiest doesn’t mean that you will as well. So, in my class, there is no one right way of doing an integral. You may use any integration technique that I’ve taught you in this class or you learned in Calculus I to evaluate integrals in this class. In other words, always take the approach that you find to be the easiest.

Note that this final point is more geared towards my class and it’s completely possible that your instructor may not agree with this and so be careful in applying this point if you aren’t in my class.

Okay, let’s get on with the strategy.

We used this idea when we were looking at integrals involving trig functions. For example, consider the following integral.

This integral can’t be done as is, however simply by recalling the identity,

the integral becomes very easy to do.

Note that this example also shows that simplification does not necessarily mean that we’ll write the integrand in a “simpler” form. It only means that we’ll write the integrand into a form that we can deal with and this is often longer and/or “messier” than the original integral.

The first integral can be done with partial fractions and the second could be done with a trig substitution.

However, both could also be evaluated using the substitution \(u = {x^2} - 1\) and the work involved in the substitution would be significantly less than the work involved in either partial fractions or trig substitution.

So, always look for quick, simple substitutions before moving on to the more complicated Calculus II techniques.

typical example here is the following integral.

This integral doesn’t obviously fit into any of the forms we looked at in this chapter. However, with the substitution \(u = \sin x\) we can reduce the integral to the form,

which is a trig substitution problem.

Don’t ever get locked into the idea that an integral will only require one step to completely evaluate it. Many will require more than one step.

As noted above this strategy is not a hard and fast set of rules. It is only intended to guide you through the process of best determining how to do any given integral. Note as well that the only place Calculus II actually arises is in the third step. Steps 1, 2 and 4 involve nothing more than manipulation of the integrand either through direct manipulation of the integrand or by using a substitution. The last two steps are simply ideas to think about in going through this strategy.

Many students go through this process and concentrate almost exclusively on Step 3 (after all this is Calculus II, so it’s easy to see why they might do that….) to the exclusion of the other steps. One very large consequence of that exclusion is that often a simple manipulation or substitution is overlooked that could make the integral very easy to do. Before moving on to the next section we should work a couple of quick problems illustrating a couple of not so obvious simplifications/manipulations and a not so obvious substitution.

This integral almost falls into the form given in 3c . It is a quotient of tangent and secant and we know that sometimes we can use the same methods for products of tangents and secants on quotients.

The process from that section tells us that if we have even powers of secant to strip two of them off and convert the rest to tangents. That won’t work here. We can split two secants off, but they would be in the denominator and they won’t do us any good there. Remember that the point of splitting them off is so they would be there for the substitution \(u = \tan x\). That requires them to be in the numerator. So, that won’t work and so we’ll have to find another solution method.

There are in fact two solution methods to this integral depending on how you want to go about it. We’ll take a look at both.

Solution 1 In this solution method we could just convert everything to sines and cosines and see if that gives us an integral we can deal with.

Note that just converting to sines and cosines won’t always work and if it does it won’t always work this nicely. Often there will be a lot more work that would need to be done to complete the integral.

Solution 2 This solution method goes back to dealing with secants and tangents. Let’s notice that if we had a secant in the numerator we could just use \(u = \sec x\) as a substitution and it would be a fairly quick and simple substitution to use. We don’t have a secant in the numerator. However, we could very easily get a secant in the numerator simply by multiplying the numerator and denominator by secant.

In the previous example we saw two “simplifications” that allowed us to do the integral. The first was using identities to rewrite the integral into terms we could deal with and the second involved multiplying the numerator and the denominator by something to again put the integral into terms we could deal with.

Using identities to rewrite an integral is an important “simplification” and we should not forget about it. Integrals can often be greatly simplified or at least put into a form that can be dealt with by using an identity.

The second “simplification” is not used as often, but does show up on occasion so again, it’s best to not forget about it. In fact, let’s take another look at an example in which multiplying the numerator and denominator by something will allow us to do an integral.

This is an integral in which if we just concentrate on the third step we won’t get anywhere. This integral doesn’t appear to be any of the kinds of integrals that we worked in this chapter.

We can do the integral however, if we do the following,

This does not appear to have done anything for us. However, if we now remember the first “simplification” we looked at above we will notice that we can use an identity to rewrite the denominator. Once we do that we can further reduce the integral into something we can deal with.

So, we’ve seen once again that multiplying the numerator and denominator by something can put the integral into a form that we can integrate. Notice as well that this example also showed that “simplifications” do not necessarily put an integral into a simpler form. They only put the integral into a form that is easier to integrate.

Let’s now take a quick look at an example of a substitution that is not so obvious.

We introduced this example saying that the substitution was not so obvious. However, this is really an integral that falls into the form given by 3e in our strategy above. However, many people miss that form and so don’t think about it. So, let’s try the following substitution.

With this substitution the integral becomes,

This is now an integration by parts integral. Remember that often we will need to use more than one technique to completely do the integral. This is a fairly simple integration by parts problem so we’ll leave the remainder of the details to you to check.

Before leaving this section we should also point out that there are integrals out there in the world that just can’t be done in terms of functions that we know. Some examples of these are.

That doesn’t mean that these integrals can’t be done at some level. If you go to a computer algebra system such as Maple or Mathematica and have it do these integrals it will return the following.

So, it appears that these integrals can in fact be done. However, this is a little misleading. Here are the definitions of each of the functions given above.

Error Function

The Sine Integral

The Fresnel Cosine Integral

The Cosine Integral

Where \(\gamma \) is the Euler-Mascheroni constant .

Note that the first three are simply defined in terms of themselves and so when we say we can integrate them all we are really doing is renaming the integral. The fourth one is a little different and yet it is still defined in terms of an integral that can’t be done in practice.

It will be possible to integrate every integral given in this class, but it is important to note that there are integrals that just can’t be done. We should also note that after we look at Series we will be able to write down series representations of each of the integrals above.

Collapse menu

Introduction, 1 analytic geometry.

2 Instantaneous Rate of Change: The Derivative

3 Rules for Finding Derivatives

4 Transcendental Functions

5 Curve Sketching

6 Applications of the Derivative

7 Integration

8 Techniques of Integration

1. substitution, 2. powers of sine and cosine, 3. trigonometric substitutions, 4. integration by parts, 5. rational functions, 6. numerical integration, 7. additional exercises, 9 applications of integration.

10 Polar Coordinates, Parametric Equations

11 Sequences and Series

12 Three Dimensions

13 Vector Functions

14 Partial Differentiation

15 Multiple Integration

16 Vector Calculus

17 Differential Equations

18 Useful formulas

19 introduction to sage.

how to solve integration problems easily

Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with $$\int x^{10}\,dx$$ we realize immediately that the derivative of $\ds x^{11}$ will supply an $\ds x^{10}$: $\ds (x^{11})'=11x^{10}$. We don't want the "11'', but constants are easy to alter, because differentiation "ignores'' them in certain circumstances, so $${d\over dx}{1\over 11}{x^{11}}={1\over 11}11{x^{10}}=x^{10}.$$

From our knowledge of derivatives, we can immediately write down a number of antiderivatives. Here is a list of those most often used:

$$\displaylines{ \int x^n\,dx={x^{n+1}\over n+1}+C, \quad\hbox{if $n\not=-1$}\cr \int x^{-1}\,dx = \ln |x|+C\cr \int e^x\,dx = e^x+C\cr \int \sin x\,dx = -\cos x+C\cr \int \cos x\,dx = \sin x+C\cr \int \sec^2 x\,dx = \tan x+C\cr \int \sec x\tan x\,dx = \sec x+C\cr \int {1\over1+x^2}\,dx = \arctan x+C\cr \int {1\over \sqrt{1-x^2}}\,dx = \arcsin x+C\cr }$$

Download on App Store

Download on App Store


This tutorial begins with a discussion of antiderivatives, mathematical objects that are closely related to derivatives. After the integral is introduced via the area problem, the integral and the antiderivative are shown to be related by an amazing theorem called the fundamental theorem of calculus. After establishing some techniques for evaluating integrals, we exhibit the important interpretation of the integral as a limit of a certain sum and demonstrate a variety of applications of the integral to problems in business and economics, geometry, and science.


how to solve integration problems easily

Suppose that during the initial stages of production the marginal cost to produce a commodity is C '(x) = 2x dollars per unit. This time, suppose the manufacturer also knows that the fixed cost of production, C(0), is $500. Find the corresponding cost function C (x). We have already seen that any cost function for this marginal cost must be of the form C(x) = x 2 + a for some constant a. Since

C (0) = 500 = 0 2 + a = a,

we have a = 500. Thus, the cost function is given by C(x) = x 2 + 500

From this example, we see that the arbitrary constant c is the fixed cost of production. Knowing only the marginal cost cannot tell us what that fixed cost is; the fixed cost is additional information. Each of the cost functions corresponding to a marginal cost of C'(x) = 2x will have the form

how to solve integration problems easily

To verify Rule (3), recall that

how to solve integration problems easily

Example 5 Use Rule (2) to evaluate each antiderivative:

how to solve integration problems easily

Math Topics

More solvers.


  1. How to solve Integration problems easily (Part 01).

    how to solve integration problems easily

  2. Reddit

    how to solve integration problems easily

  3. How to find Integration Problems Easily|How to Solve Integration Problems|Integration By Parts

    how to solve integration problems easily

  4. how to solve integration questions easily||indefinite integration tricks||Integration problems

    how to solve integration problems easily

  5. Solve Life Problems with 3 Effective Steps Strategy by Shruti Vashisht

    how to solve integration problems easily

  6. How to Solve Integral Calculus Problems Easily

    how to solve integration problems easily


  1. Solve the integral problem

  2. Solution: Definite Integrals Without Calculus


  4. Integration Solved problems 1 to 7//INTERMEDIATE IPE,CBSE,ISC


  6. O Level


  1. ❖ Basic Integration Problems

    Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) !

  2. Integration Tricks (That Teachers Won't Tell You) for Integral Calculus

    Here are some amazing tricks for solving integrals. I promise your calculus teacher won't show you these integration tricks.

  3. Solve Integration Problems in 60 Seconds, Shortcut ...


  4. What are the best techniques in solving integration problems easily?

    1.) Make sure you have a solid understanding of differential calculus. 2.) Familiarize yourself with summation notation and respective formulas. 3.) Make sure

  5. How to Solve Integration Problems

    How does one solve an integral problem? ... Integrals are solved various ways depending on the function being evaluated. The most basic way is to

  6. Integration Tricks

    Many challenging integration problems can be solved surprisingly quickly by simply knowing the right technique to apply. While finding the right technique

  7. Some “Tricks” for Integration

    Math107. Fall 2007. Calculus II. University of Nebraska-Lincoln. Some “Tricks” for Integration. Trick. Examples. Expand. ∫. (1 + ex).

  8. Calculus II

    1. General Tips · 2. Taking Notes · 3. Getting Help · 4. Doing Homework · 5. Problem Solving · 6. Studying For an Exam · 7. Taking an Exam · 8. Learn

  9. 8. Techniques of Integration

    Sometimes this is a simple problem, since it will be apparent that the function you wish to ... We don't want the "11'', but constants are easy to alter

  10. Integrate a function with Step-by-Step Math Problem Solver

    Integrate any expression with our integration solver. ... All such problems are solved by antidifferentiation. An elementary example from business is the