Wize University Calculus 2 Textbook > Integration Techniques

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
We have a product of trigonometric functions

Strategy

You will often need to rewrite the integral using trig identities

Wize Concept
Basic identities‾tan⁡x=sin⁡xcos⁡xsin⁡2x+cos⁡2x=1tan⁡2x+1=sec⁡2xReciprocals‾sec⁡x=1cos⁡xcsc⁡x=1sin⁡xcot⁡x=1tan⁡xDouble & half angle identities‾sin⁡(2x)=2sin⁡xcos⁡xcos⁡2x=1+cos⁡(2x)2sin⁡2x=1−cos⁡(2x)2Angle sum & difference identities‾sin⁡A cos⁡B=12[sin⁡(A−B)+sin⁡(A+B)]sin⁡A sin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]cos⁡A cos⁡B=12[cos⁡(A−B)+cos⁡(A+B)]\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}Basic identities​tanx=cosxsinx​sin2x+cos2x=1tan2x+1=sec2xReciprocals​secx=cosx1​cscx=sinx1​cotx=tanx1​Double & half angle identities​sin(2x)=2sinxcosxcos2x=21+cos(2x)​sin2x=21−cos(2x)​Angle sum & difference identities​sinA cosB=21​[sin(A−B)+sin(A+B)]sinA sinB=21​[cos(A−B)−cos(A+B)]cosA cosB=21​[cos(A−B)+cos(A+B)]​​


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

Wize Concept
ddx(sin⁡x)=cos⁡xddx(cos⁡x)=−sin⁡xddx(tan⁡x)=sec⁡2xddx(sec⁡x)=sec⁡xtan⁡xddx(csc⁡x)=−csc⁡xcot⁡xddx(cot⁡x)=−csc⁡2x\begin{array}{l}

\color{black}\frac{d}{dx}\left(\sin x\right)=\cos x\\\\
\color{black}\frac{d}{dx}\left(\cos x\right)=-\sin x\\\\
\color{black}\frac{d}{dx}\left(\tan x\right)=\sec^2 x\\\\
\color{black}\frac{d}{dx}\left(\sec x\right)=\sec x\tan x\\\\
\color{black}\frac{d}{dx}\left(\csc x\right)=-\csc x\cot x\\\\
\color{black}\frac{d}{dx}\left(\cot x\right)=-\csc^2 x\\\\
\end{array}dxd​(sinx)=cosxdxd​(cosx)=−sinxdxd​(tanx)=sec2xdxd​(secx)=secxtanxdxd​(cscx)=−cscxcotxdxd​(cotx)=−csc2x​

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Some note-worthy special trig integrals
∫tan⁡x dx=−ln⁡cos⁡x+C
\color{orange}\displaystyle\int_{ }^{ }\tan x\ dx=-\ln\cos x+C∫​tanx dx=−lncosx+C
Rewrite tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}tanx=cosxsinx​
Then use a u-substitution *(Try this as an exercise)

∫sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C\color{orange}\displaystyle\int_{ }^{ }\sec x\ dx=\ln\left|\sec x+\tan x\right|+C∫​secx dx=ln∣secx+tanx∣+C
Multiply by sec⁡x+tan⁡xsec⁡x+tan⁡x\frac{\sec x+\tan x}{\sec x+\tan x}secx+tanxsecx+tanx​  
Then use a u-substitution *(See Concept Clarifier)

∫csc⁡x dx=−ln⁡∣csc⁡x+cot⁡x∣+C\color{orange}\displaystyle\int_{ }^{ }\csc x\ dx=-\ln\left|\csc x+\cot x\right|+C∫​cscx dx=−ln∣cscx+cotx∣+C
Multiply by csc⁡x+cot⁡xcsc⁡x+cot⁡x\frac{\csc x+\cot x}{\csc x+\cot x}cscx+cotxcscx+cotx​  
Then use a u-substitution *(Similar to ∫sec⁡x dx\int_{ }^{ }\sec x\ dx∫​secx dx)

∫sec⁡3x dx=sec⁡x tan⁡x+ln⁡∣sec⁡x+tan⁡x∣2+C\color{orange}\displaystyle\int_{ }^{ }\sec^3x\ dx=\frac{\sec x\ \tan x+\ln\left|\sec x+\tan x\right|}{2}+C∫​sec3x dx=2secx tanx+ln∣secx+tanx∣​+C
Rewrite as ∫sec⁡x (sec⁡2x)dx\int_{ }^{ }\sec x\ \left(\sec^2x\right)dx∫​secx (sec2x)dx
Then use Integration by Parts (See practice questions)

Practice Question
Find ∫sin⁡4xcos⁡3x   ⁣dx{\displaystyle\int}\sin^4x\cos^3x\space\de{x}∫sin4xcos3x  dx.

Hint: 
Factor out one copy of cos⁡x\cos xcosx
Convert the rest to sin⁡x\sin xsinx using the identity cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2xcos2x=1−sin2x
Use u-substitution

 sin⁡5x5−sin⁡7x7+C\frac{\sin^5x}{5}-\frac{\sin^7x}{7}+C5sin5x​−7sin7x​+C

 sin⁡5x5−sin⁡6x6+C\frac{\sin^5x}{5}-\frac{\sin^6x}{6}+C5sin5x​−6sin6x​+C

 sin⁡5x5sin⁡4x4+C\frac{\sin^5x}{5}\frac{\sin^4x}{4}+C5sin5x​4sin4x​+C

 sin⁡5x5cos⁡4x4+C\frac{\sin^5x}{5}\frac{\cos^4x}{4}+C5sin5x​4cos4x​+C

 None of the above

I don't know

Practice Question
Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.
Hint: 
Factor out one copy of sin⁡x\sin xsinx
Convert the rest to cos⁡x\cos xcosx using the identity sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2xsin2x=1−cos2x
Use u-substitution

 sin⁡8x8cos⁡x+C\frac{\sin^8x}{8\cos x}+C8cosxsin8x​+C

 sin⁡8x8+C\frac{\sin^8x}{8}+C8sin8x​+C

 sin⁡8x cos⁡2x16 +C\frac{\sin^8x\ \cos^2x}{16}\ +C16sin8x cos2x​ +C

 cos⁡3x−3cos⁡5x5+cos⁡7x7+C\cos^3x-\frac{3\cos^5x}{5}+\frac{\cos^7x}{7}+Ccos3x−53cos5x​+7cos7x​+C

 −cos⁡x+cos⁡3x−3cos⁡5x5+cos⁡7x7+C-\cos x+\cos^3x-\frac{3\cos^5x}{5}+\frac{\cos^7x}{7}+C−cosx+cos3x−53cos5x​+7cos7x​+C

I don't know

Practice Question
Evaluate ∫sin⁡2xcos⁡2x   ⁣dx{\displaystyle\int} \sin^2x\cos^2x\space\de{x}∫sin2xcos2x  dx.
Hint: 
Use the half angle formulas sin⁡2x=1−cos⁡2x2\sin^2x=\frac{1-\cos2x}{2}sin2x=21−cos2x​ and cos⁡2x=1+cos⁡2x2\cos^2x=\frac{1+\cos2x}{2}cos2x=21+cos2x​

x − sin⁡(4x)4+Cx\ -\ \frac{\sin(4x)}{4}+Cx − 4sin(4x)​+C

x2 − sin⁡(4x)8+C\frac{x}{2}\ -\ \frac{\sin(4x)}{8}+C2x​ − 8sin(4x)​+C

x8 − sin⁡(4x)32+C\frac{x}{8}\ -\ \frac{\sin(4x)}{32}+C8x​ − 32sin(4x)​+C

sin⁡3xcos⁡2x3+sin⁡2xcos⁡3x3+C\frac{\sin^3x\cos^2x}{3}+\frac{\sin^2x\cos^3x}{3}+C3sin3xcos2x​+3sin2xcos3x​+C

sin⁡3xcos⁡2x3−sin⁡2xcos⁡3x3+C\frac{\sin^3x\cos^2x}{3}-\frac{\sin^2x\cos^3x}{3}+C3sin3xcos2x​−3sin2xcos3x​+C

I don't know

Practice Question
Find∫sec⁡8xtan⁡6x   ⁣dx{\displaystyle\int}\sec^8x\tan^6x\space\de{x}∫sec8xtan6x  dx.
Hint: 
Factor out one copy of sec⁡2x\sec^2xsec2x
Convert the rest to tan⁡x\tan xtanx using the identity sec⁡2x=1+tan⁡2x\sec^2x=1+\tan^2xsec2x=1+tan2x
Use u-substitution

tan⁡13x13+3tan⁡11x11+tan⁡9x3+tan⁡7x7+C\frac{\tan^{13}x}{13}+\frac{3\tan^{11}x}{11}+\frac{\tan^9x}{3}+\frac{\tan^7x}{7}+C13tan13x​+113tan11x​+3tan9x​+7tan7x​+C

tan⁡13x13+5tan⁡11x11+5tan⁡9x9+tan⁡7x7+C\frac{\tan^{13}x}{13}+\frac{5\tan^{11}x}{11}+\frac{5\tan^9x}{9}+\frac{\tan^7x}{7}+C13tan13x​+115tan11x​+95tan9x​+7tan7x​+C

sec⁡9xtan⁡6x9+sec⁡8xtan⁡7x7+C\frac{\sec^9x\tan^6x}{9}+\frac{\sec^8x\tan^7x}{7}+C9sec9xtan6x​+7sec8xtan7x​+C

sec⁡9xtan⁡7x9+sec⁡8xtan⁡7x7+C\frac{\sec^9x\tan^7x}{9}+\frac{\sec^8x\tan^7x}{7}+C9sec9xtan7x​+7sec8xtan7x​+C

sec⁡9xtan⁡6x9−sec⁡8xtan⁡7x7+C\frac{\sec^9x\tan^6x}{9}-\frac{\sec^8x\tan^7x}{7}+C9sec9xtan6x​−7sec8xtan7x​+C

I don't know

Practice Question
Evaluate ∫−π0sec⁡3xtan⁡3xdx\displaystyle \int_{ -\pi}^{0 }\sec^3x\tan^3xdx∫−π0​sec3xtan3xdx.
Hint:
Factor out sec⁡xtan⁡x\sec x\tan xsecxtanx
Convert the rest of the tan⁡x\tan xtanx into sec⁡x\sec xsecx using tan⁡2x=sec⁡2x−1\tan^2x=\sec^2x-1tan2x=sec2x−1
Use a u-sub

Practice Question
Evaluate ∫sin⁡(3x)sin⁡(2x)   ⁣dx{\displaystyle \int} \sin(3x)\sin(2x)\ \de{x}∫sin(3x)sin(2x)  dx.
Hint: 
Use the angle sum and difference formula sin⁡A sin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A\ \sin B=\frac{1}{2}\left[\cos\left(A-B\right)-\cos\left(A+B\right)\right]sinA sinB=21​[cos(A−B)−cos(A+B)]

 sin⁡x2−sin⁡5x10+C\frac{\sin x}{2}-\frac{\sin5x}{10}+C2sinx​−10sin5x​+C

 sin⁡x−sin⁡5x5+C\sin x-\frac{\sin5x}{5}+Csinx−5sin5x​+C

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

 cos⁡(3x)cos⁡(2x)6+C\frac{\cos\left(3x\right)\cos\left(2x\right)}{6}+C6cos(3x)cos(2x)​+C

None of the above

I don't know

Example: Special Case
Find ∫sec⁡x dx\int_{ }^{ }\sec x\ dx∫​secx dx.

Hint:
Multiply by sec⁡x+tan⁡xsec⁡x+tan⁡x\frac{\sec x+\tan x}{\sec x+\tan x}secx+tanxsecx+tanx​
Then use a u-substitution

 tan⁡2x2+C\frac{\tan^2x}{2}+C2tan2x​+C

 ln⁡∣sec⁡x+tan⁡x∣+C\ln\left|\sec x+\tan x\right|+Cln∣secx+tanx∣+C

 12(sec⁡x+tan⁡x)2+C\frac{1}{2\left(\sec x+\tan x\right)^2}+C2(secx+tanx)21​+C

 −2(sec⁡x+tan⁡x)2+C-\frac{2}{\left(\sec x+\tan x\right)^2}+C−(secx+tanx)22​+C

None of the above

I don't know

Practice Question
Evaluate∫sec⁡3x   ⁣dx{\displaystyle\int}\sec^3x \space\de{x}∫sec3x  dx.
Hint:
Rewrite as ∫sec⁡x (sec⁡2x)dx\int_{ }^{ }\sec x\ \left(\sec^2x\right)dx∫​secx (sec2x)dx
Then use integration by parts with u=sec⁡x,   dv=sec⁡2x dxu=\sec x,\ \ \ dv=\sec^2x\ dxu=secx,   dv=sec2x dx

 12ln⁡∣sec⁡x+tan⁡x∣+C\space\frac{1}{2}\ln\vert\sec x+\tan x\vert+C 21​ln∣secx+tanx∣+C

sec⁡xtan⁡x2 − 12ln⁡∣sec⁡x−tan⁡x∣+C\frac{\sec x\tan x}{2}\space-\space\frac{1}{2}\ln\vert\sec x-\tan x\vert+C2secxtanx​ − 21​ln∣secx−tanx∣+C

sec⁡4x4tan⁡x−12ln⁡∣sec⁡x−tan⁡x∣+C\frac{\sec^4x}{4\tan x}-\frac{1}{2}\ln\left|\sec x-\tan x\right|+C4tanxsec4x​−21​ln∣secx−tanx∣+C

sec⁡xtan⁡x2 + 12ln⁡∣sec⁡x+tan⁡x∣+C\frac{\sec x\tan x}{2}\space+\space\frac{1}{2}\ln\vert\sec x+\tan x\vert+C2secxtanx​ + 21​ln∣secx+tanx∣+C

sec⁡4x4tan⁡x+C\frac{\sec^4x}{4\tan x}+C4tanxsec4x​+C

I don't know

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

Know these!
1. In=∫(tan⁡x)n dx\displaystyle I_n=\int_{ }^{ }\left(\tan x\right)^n\ dxIn​=∫​(tanx)n dx   →   In=(tan⁡x)n−1n−1−In−2\displaystyle I_n=\frac{\left(\tan x\right)^{n-1}}{n-1}-I_{n-2}In​=n−1(tanx)n−1​−In−2​
Use trig integrals and take out a factor of tan⁡2x\tan^2xtan2x


2. In=∫(sin⁡x)n dx\displaystyle I_n=\int\left(\sin x\right)^n\ dxIn​=∫(sinx)n dx   →   In=−cos⁡x(sin⁡x)n−1n+n−1nIn−2\displaystyle I_n=-\frac{\cos x\left(\sin x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}In​=−ncosx(sinx)n−1​+nn−1​In−2​
Use integration by parts


3. In=∫(cos⁡x)n dx\displaystyle I_n=\int\left(\cos x\right)^n\ dxIn​=∫(cosx)n dx  →  In=(cos⁡x)n−1sin⁡xn+n−1nIn−2\displaystyle 
I_n=\frac{\left(\cos x\right)^{n-1}\sin x}{n}+\frac{n-1}{n}I_{n-2}In​=n(cosx)n−1sinx​+nn−1​In−2​
Use integration by parts

Practice Question
Given that In=∫(sin⁡x)n dx\displaystyle I_n=\int\left(\sin x\right)^n\ dxIn​=∫(sinx)n dx, find the reduction formula of the form In=f(x)+n−1nIn−2\displaystyle I_n=f(x)+\frac{n-1}{n}I_{n-2}In​=f(x)+nn−1​In−2​, for n≥2n\ge2n≥2.

 In=−cos⁡x(sin⁡x)n−1n+n−1nIn−2\displaystyle I_n=-\frac{\cos x\left(\sin x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}In​=−ncosx(sinx)n−1​+nn−1​In−2​

 In=cos⁡x(sin⁡x)n−1n+n−1nIn−2\displaystyle I_n=\frac{\cos x\left(\sin x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}In​=ncosx(sinx)n−1​+nn−1​In−2​

 In=sin⁡x(cos⁡x)n−1n+n−1nIn−2\displaystyle I_n=\frac{\sin x\left(\cos x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}In​=nsinx(cosx)n−1​+nn−1​In−2​

 In=−sin⁡x(cos⁡x)n−1n+n−1nIn−2\displaystyle I_n=-\frac{\sin x\left(\cos x\right)^{n-1}}{n}+\frac{n-1}{n}I_{n-2}In​=−nsinx(cosx)n−1​+nn−1​In−2​

 None of the above

I don't know

Practice: Reduction Formula
Given that In=∫(tan⁡x)ndx\displaystyle I_n=\int (\tan x)^ndxIn​=∫(tanx)ndx, find the reduction formula of the form In=f(x)−In−2I_n=f(x)-I_{n-2}In​=f(x)−In−2​, for n≥2n\ge2n≥2.

In=(tan⁡x)n−1n−In−2\displaystyle I_n=\frac{\left(\tan x\right)^{n-1}}{n}-I_{n-2}In​=n(tanx)n−1​−In−2​

In=(tan⁡x)n−1n−1−In−2\displaystyle I_n=\frac{\left(\tan x\right)^{n-1}}{n-1}-I_{n-2}In​=n−1(tanx)n−1​−In−2​

In=(tan⁡x)nn−In−2\displaystyle I_n=\frac{\left(\tan x\right)^{n}}{n}-I_{n-2}In​=n(tanx)n​−In−2​

In=(tan⁡x)n+1n+1−In−2\displaystyle I_n=\frac{\left(\tan x\right)^{n+1}}{n+1}-I_{n-2}In​=n+1(tanx)n+1​−In−2​

In=−(tan⁡x)n−1n−In−2\displaystyle I_n=-\frac{\left(\tan x\right)^{n-1}}{n}-I_{n-2}In​=−n(tanx)n−1​−In−2​

I don't know

Extra Practice

Trigonometric Integral

Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.

Trigonometric Integral

Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.

Trigonometric Integral

Practice Question
Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.

Trigonometric Integral

Practice Question
Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.

Trigonometric Integral

Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.

Practice: Trigonometric Integral, with even powers

Practice Question
Find ∫cos⁡4(x)  ⁣dx{\displaystyle \int}\cos^4(x)\de{x}∫cos4(x) dx.

Trigonometric Integral

Practice Question
Compute ∫sin⁡7x  ⁣dx{\displaystyle\int}\sin^7x\de{x}∫sin7x dx.