Wize University Calculus 2 Textbook > Integration Techniques

Trig Integrals

Trigonometric Integrals (Disclaimer: Please note that the reciprocal of cscx should read 1/sinx and not 1/secx as indicated within the attached video).

Identifying Clues

We have a product of trigonometric functions

Strategy

You will often need to rewrite the integral using trig identities

Wize Concept

$\begin{array}{l} \displaystyle \color{black}\underline{\text{Basic identities}}\\ \color{black}\tan x=\frac{\sin x}{\cos x}\\ \color{black}\sin^2x+\cos^2x=1\\ \color{black}\tan^2x+1=\sec^2x\\\\\hline \color{black}\underline{\text{Reciprocals}}\\ \color{black}\sec x=\frac{1}{\cos x}\\ \color{black}\csc x=\frac{1}{\sin x}\\ \color{black}\cot x=\frac{1}{\tan x}\\\\\hline \color{black}\underline{\text{Double \& half angle identities}}\\ \color{black}\sin(2x)=2\sin x\cos x\\ \color{black}\cos^2x=\frac{1+\cos(2x)}{2}\\ \color{black}\sin^2x=\frac{1-\cos(2x)}{2}\\\\\hline \color{black}\underline{\text{Angle sum \& difference identities}}\\ \color{black}\sin A\ \cos B=\frac{1}{2}\left[\sin\left(A-\color{black}B\right)+\sin\left(A+B\right)\right]\\ \color{black}\sin A\ \sin B=\frac{1}{2}\left[\cos\left(A-B\right)-\color{black}\cos\left(A+B\right)\right]\\ \color{black}\cos A\ \cos B=\frac{1}{2}\left[\cos\left(A-\color{black}B\right)+\cos\left(A+B\right)\right]\\\\ \end{array}$

Try u-substitution: let $u$ be one part of the function such that $u'$ shows up elsewhere in the expression

Some note-worthy special trig integrals

$\color{orange}\displaystyle\int_{ }^{ }\tan x\ dx=-\ln\cos x+C$

Rewrite $\tan x=\frac{\sin x}{\cos x}$
Then use a u-substitution *(Try this as an exercise)

$\color{orange}\displaystyle\int_{ }^{ }\sec x\ dx=\ln\left|\sec x+\tan x\right|+C$

Multiply by $\frac{\sec x+\tan x}{\sec x+\tan x}$
Then use a u-substitution *(See Concept Clarifier)

$\color{orange}\displaystyle\int_{ }^{ }\csc x\ dx=-\ln\left|\csc x+\cot x\right|+C$

Multiply by $\frac{\csc x+\cot x}{\csc x+\cot x}$
Then use a u-substitution *(Similar to $\int_{ }^{ }\sec x\ dx$ )

$\color{orange}\displaystyle\int_{ }^{ }\sec^3x\ dx=\frac{\sec x\ \tan x+\ln\left|\sec x+\tan x\right|}{2}+C$

Rewrite as $\int_{ }^{ }\sec x\ \left(\sec^2x\right)dx$
Then use Integration by Parts (See practice questions)

Practice Question

Find ${\displaystyle\int}\sin^4x\cos^3x\space\de{x}$ .

Hint:

Factor out one copy of $\cos x$
Convert the rest to $\sin x$ using the identity $\cos^2x=1-\sin^2x$
Use u-substitution

$\frac{\sin^5x}{5}-\frac{\sin^7x}{7}+C$

$\frac{\sin^5x}{5}-\frac{\sin^6x}{6}+C$

$\frac{\sin^5x}{5}\frac{\sin^4x}{4}+C$

$\frac{\sin^5x}{5}\frac{\cos^4x}{4}+C$

None of the above

I don't know

Practice Question

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Hint:

Factor out one copy of $\sin x$
Convert the rest to $\cos x$ using the identity $\sin^2x=1-\cos^2x$
Use u-substitution

$\frac{\sin^8x}{8\cos x}+C$

$\frac{\sin^8x}{8}+C$

$\frac{\sin^8x\ \cos^2x}{16}\ +C$

$\cos^3x-\frac{3\cos^5x}{5}+\frac{\cos^7x}{7}+C$

$-\cos x+\cos^3x-\frac{3\cos^5x}{5}+\frac{\cos^7x}{7}+C$

I don't know

Practice Question

Evaluate ${\displaystyle\int} \sin^2x\cos^2x\space\de{x}$ .

Hint:

Use the half angle formulas $\sin^2x=\frac{1-\cos2x}{2}$ and $\cos^2x=\frac{1+\cos2x}{2}$

$x\ -\ \frac{\sin(4x)}{4}+C$

$\frac{x}{2}\ -\ \frac{\sin(4x)}{8}+C$

$\frac{x}{8}\ -\ \frac{\sin(4x)}{32}+C$

$\frac{\sin^3x\cos^2x}{3}+\frac{\sin^2x\cos^3x}{3}+C$

$\frac{\sin^3x\cos^2x}{3}-\frac{\sin^2x\cos^3x}{3}+C$

I don't know

Practice Question

Find ${\displaystyle\int}\sec^8x\tan^6x\space\de{x}$ .

Hint:

Factor out one copy of $\sec^2x$
Convert the rest to $\tan x$ using the identity $\sec^2x=1+\tan^2x$
Use u-substitution

$\frac{\tan^{13}x}{13}+\frac{3\tan^{11}x}{11}+\frac{\tan^9x}{3}+\frac{\tan^7x}{7}+C$

$\frac{\tan^{13}x}{13}+\frac{5\tan^{11}x}{11}+\frac{5\tan^9x}{9}+\frac{\tan^7x}{7}+C$

$\frac{\sec^9x\tan^6x}{9}+\frac{\sec^8x\tan^7x}{7}+C$

$\frac{\sec^9x\tan^7x}{9}+\frac{\sec^8x\tan^7x}{7}+C$

$\frac{\sec^9x\tan^6x}{9}-\frac{\sec^8x\tan^7x}{7}+C$

I don't know

Practice Question

Evaluate $\displaystyle \int_{ -\pi}^{0 }\sec^3x\tan^3xdx$ .

Hint:

Factor out $\sec x\tan x$
Convert the rest of the $\tan x$ into $\sec x$ using $\tan^2x=\sec^2x-1$
Use a u-sub

I don't know

Practice Question

Evaluate $\int} \sin(3x)\sin(2x)\ \de{x$ .

Hint:

Use the angle sum and difference formula $\sin A\ \sin B=\frac{1}{2}\left[\cos\left(A-B\right)-\cos\left(A+B\right)\right]$

$\frac{\sin x}{2}-\frac{\sin5x}{10}+C$

$\sin x-\frac{\sin5x}{5}+C$

$-\frac{\cos\left(3x\right)\cos\left(2x\right)}{6}+C$

$\frac{\cos\left(3x\right)\cos\left(2x\right)}{6}+C$

None of the above

I don't know

Example: Special Case

Find $\int_{ }^{ }\sec x\ dx$ .

Hint:

Multiply by $\frac{\sec x+\tan x}{\sec x+\tan x}$
Then use a u-substitution

$\frac{\tan^2x}{2}+C$

$\ln\left|\sec x+\tan x\right|+C$

$\frac{1}{2\left(\sec x+\tan x\right)^2}+C$

$-\frac{2}{\left(\sec x+\tan x\right)^2}+C$

None of the above

I don't know

Practice Question

Evaluate ${\displaystyle\int}\sec^3x \space\de{x}$ .

Hint:

Rewrite as $\int_{ }^{ }\sec x\ \left(\sec^2x\right)dx$
Then use integration by parts with $u=\sec x,\ \ \ dv=\sec^2x\ dx$

$\space\frac{1}{2}\ln\vert\sec x+\tan x\vert+C$

$\frac{\sec x\tan x}{2}\space-\space\frac{1}{2}\ln\vert\sec x-\tan x\vert+C$

$\frac{\sec^4x}{4\tan x}-\frac{1}{2}\ln\left|\sec x-\tan x\right|+C$

$\frac{\sec x\tan x}{2}\space+\space\frac{1}{2}\ln\vert\sec x+\tan x\vert+C$

$\frac{\sec^4x}{4\tan x}+C$

I don't know

Reduction Formula

When the integral takes on the form $\displaystyle I_n=\int \text{trig}^nx\ dx$ , we can integrate it using trig integrals to end up with the form $\displaystyle I_n=f(x)+[...]I_{n-2}$

Know these!

1. $\displaystyle I_n=\int_{ }^{ }\left(\tan x\right)^n\ dx$ → $\displaystyle I_n=\frac{\left(\tan x\right)^{n-1}}{n-1}-I_{n-2}$

Use trig integrals and take out a factor of $\tan^2x$

2. $\displaystyle I_n=\int\left(\sin x\right)^n\ dx$ → $\displaystyle I_n=-\frac{\cos x\left(\sin x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}$

Use integration by parts

3. $\displaystyle I_n=\int\left(\cos x\right)^n\ dx$ → $\displaystyle I_n=\frac{\left(\cos x\right)^{n-1}\sin x}{n}+\frac{n-1}{n}I_{n-2}$

Use integration by parts

Practice Question

Given that $\displaystyle I_n=\int\left(\sin x\right)^n\ dx$ , find the reduction formula of the form $\displaystyle I_n=f(x)+\frac{n-1}{n}I_{n-2}$ , for $n\ge2$ .

$\displaystyle I_n=-\frac{\cos x\left(\sin x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}$

$\displaystyle I_n=\frac{\cos x\left(\sin x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}$

$\displaystyle I_n=\frac{\sin x\left(\cos x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}$

$\displaystyle I_n=-\frac{\sin x\left(\cos x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}$

None of the above

I don't know

Practice: Reduction Formula

Given that $\displaystyle I_n=\int (\tan x)^ndx$ , find the reduction formula of the form $I_n=f(x)-I_{n-2}$ , for $n\ge2$ .

$\displaystyle I_n=\frac{\left(\tan x\right)^{n-1}}{n}-I_{n-2}$

$\displaystyle I_n=\frac{\left(\tan x\right)^{n-1}}{n-1}-I_{n-2}$

$\displaystyle I_n=\frac{\left(\tan x\right)^{n}}{n}-I_{n-2}$

$\displaystyle I_n=\frac{\left(\tan x\right)^{n+1}}{n+1}-I_{n-2}$

$\displaystyle I_n=-\frac{\left(\tan x\right)^{n-1}}{n}-I_{n-2}$

I don't know

Extra Practice

Trigonometric Integral

Practice Question

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Practice: Trigonometric Integral, with even powers

Practice Question

Find $\int}\cos^4(x)\de{x$ .

Trigonometric Integral

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Trigonometric Integral

Practice Question

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Trigonometric Integral

Practice Question

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Trigonometric Integral

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Trigonometric Integral

Compute ${\displaystyle\int}\sin^7x\de{x}$ .

Trig Integrals

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Trigonometric Integrals (Disclaimer: Please note that the reciprocal of cscx should read 1/sinx and not 1/secx as indicated within the attached video).

Identifying Clues

Practice Question

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Practice Question

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Practice Question

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Practice Question

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Practice Question

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Practice Question

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Example: Special Case

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Practice Question

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Reduction Formula

Know these!

Practice Question

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Practice: Reduction Formula

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Extra Practice

Practice Question

Practice Question

Practice Question

Practice Question