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## Integration Tricks

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- Aareyan Manzoor
- Prasun Biswas
- Henry Maltby
- Christopher Williams
- Andrew Ellinor
- Michael Fuller
- Brandon Monsen
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Consider, for instance, the antiderivative

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

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

## Odd and Even Functions

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

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 \(0\). \(_\square\)

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

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

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

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

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!

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

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

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. \(_\square\)

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

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 \(I^{-1}?\)

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

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.

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. \(_\square\)

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 \[\ln\left(\tan\frac{\theta}{2}\right)+C=-\ln(\cot x+\csc x)+C=\ln\frac{\sin\theta}{\cos\theta+1}+C.\]

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

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

\[\dfrac{d}{dx} \displaystyle \int_{g(x)}^{h(x)} f(x,t)dt = f\big(x,h(x)\big)\dfrac{d}{dx}h(x) - f\big(x,g(x)\big)\dfrac{d}{dx}g(x) + \displaystyle \int_{g(x)}^{h(x)}\dfrac{\partial}{\partial x}f(x,t)dt. \]

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

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:

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

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

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 \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx = 2\pi e^0 \cos(0) = 2\pi.\ _\square\]

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## Section 7.7 : Integration Strategy

Okay, let’s get on with the strategy.

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

the integral becomes very easy to do.

- Is the integrand a rational expression ( i.e is the integrand a polynomial divided by a polynomial)? If so, then partial fractions may work on the integral.
- Is the integrand a polynomial times a trig function, exponential, or logarithm? If so, then integration by parts may work.
- Is the integrand a product of sines and cosines, secant and tangents, or cosecants and cotangents? If so, then the topics from the second section may work. Likewise, don’t forget that some quotients involving these functions can also be done using these techniques.
- Does the integrand involve \(\sqrt {{b^2}{x^2} + {a^2}} \), \(\sqrt {{b^2}{x^2} - {a^2}} \), or \(\sqrt {{a^2} - {b^2}{x^2}} \)? If so, then a trig substitution might work nicely.
- Does the integrand have roots other than those listed above in it? If so, then the substitution \(u = \sqrt[n]{{g\left( x \right)}}\) might work.
- Does the integrand have a quadratic in it? If so, then completing the square on the quadratic might put it into a form that we can deal with.

typical example here is the following integral.

which is a trig substitution problem.

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

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

With this substitution the integral becomes,

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

## Collapse menu

Introduction, 1 analytic geometry.

## 2 Instantaneous Rate of Change: The Derivative

## 3 Rules for Finding Derivatives

- 1. The Power Rule
- 2. Linearity of the Derivative
- 3. The Product Rule
- 4. The Quotient Rule
- 5. The Chain Rule

## 4 Transcendental Functions

- 1. Trigonometric Functions
- 2. The Derivative of $\sin x$
- 3. A hard limit
- 4. The Derivative of $\sin x$, continued
- 5. Derivatives of the Trigonometric Functions
- 6. Exponential and Logarithmic functions
- 7. Derivatives of the exponential and logarithmic functions
- 8. Implicit Differentiation
- 9. Inverse Trigonometric Functions
- 10. Limits revisited
- 11. Hyperbolic Functions

## 5 Curve Sketching

- 1. Maxima and Minima
- 2. The first derivative test
- 3. The second derivative test
- 4. Concavity and inflection points
- 5. Asymptotes and Other Things to Look For

## 6 Applications of the Derivative

## 7 Integration

## 8 Techniques of Integration

- 1. Area between curves
- 2. Distance, Velocity, Acceleration
- 4. Average value of a function
- 6. Center of Mass
- 7. Kinetic energy; improper integrals
- 8. Probability
- 9. Arc Length
- 10. Surface Area

## 10 Polar Coordinates, Parametric Equations

- 1. Polar Coordinates
- 2. Slopes in polar coordinates
- 3. Areas in polar coordinates
- 4. Parametric Equations
- 5. Calculus with Parametric Equations

## 11 Sequences and Series

- 1. Sequences
- 3. The Integral Test
- 4. Alternating Series
- 5. Comparison Tests
- 6. Absolute Convergence
- 7. The Ratio and Root Tests
- 8. Power Series
- 9. Calculus with Power Series
- 10. Taylor Series
- 11. Taylor's Theorem
- 12. Additional exercises

## 12 Three Dimensions

- 1. The Coordinate System
- 3. The Dot Product
- 4. The Cross Product
- 5. Lines and Planes
- 6. Other Coordinate Systems

## 13 Vector Functions

## 14 Partial Differentiation

- 1. Functions of Several Variables
- 2. Limits and Continuity
- 3. Partial Differentiation
- 4. The Chain Rule
- 5. Directional Derivatives
- 6. Higher order derivatives
- 7. Maxima and minima
- 8. Lagrange Multipliers

## 15 Multiple Integration

- 1. Volume and Average Height
- 2. Double Integrals in Cylindrical Coordinates
- 3. Moment and Center of Mass
- 4. Surface Area
- 5. Triple Integrals
- 6. Cylindrical and Spherical Coordinates
- 7. Change of Variables

## 16 Vector Calculus

- 1. Vector Fields
- 2. Line Integrals
- 3. The Fundamental Theorem of Line Integrals
- 4. Green's Theorem
- 5. Divergence and Curl
- 6. Vector Functions for Surfaces
- 7. Surface Integrals
- 8. Stokes's Theorem
- 9. The Divergence Theorem

## 17 Differential Equations

- 1. First Order Differential Equations
- 2. First Order Homogeneous Linear Equations
- 3. First Order Linear Equations
- 4. Approximation
- 5. Second Order Homogeneous Equations
- 6. Second Order Linear Equations
- 7. Second Order Linear Equations, take two

## 18 Useful formulas

- Solve equations and inequalities
- Simplify expressions
- Factor polynomials
- Graph equations and inequalities
- Advanced solvers
- All solvers
- Arithmetics
- Determinant
- Percentages
- Scientific Notation
- Inequalities

## INTEGRATION

## Antiderivatives

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

To verify Rule (3), recall that

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

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## VIDEO

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

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