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Wize AP Calculus (BC) Textbook > Integration Techniques

Trigonometric Integrals

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Trigonometric Integrals

When integrating complex trigonometric functions, we often need to use a combination of substitutions and trig identities.

Case 1: ∫sin⁡mxcos⁡nx  ⁣dx{\displaystyle\int} \sin^m x \cos^n x \de{x}

  • Power of sin⁡x\sin x is odd
  • Keep a copy of sin⁡x\sin x and convert the rest to cos⁡x\cos x using sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x
  • Formula: ∫cos⁡nx(1−cos⁡2x)ksin⁡x  ⁣dx{\displaystyle \int}\cos^nx(1-\cos^2x)^k\sin x\de{x}
  • Do a U-substitution with u=cos⁡xu=\cos x
  • Power of cos⁡x\cos x is odd
  • Keep a copy of cos⁡x\cos x and convert the rest to sin⁡x\sin x using cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x
  • Formula: ∫sin⁡mx(1−sin⁡2x)kcos⁡x  ⁣dx{\displaystyle \int}\sin^mx(1-\sin^2x)^k\cos x\de{x}
  • Do a U-substitution with u=sin⁡xu=\sin x
  • Both powers of sin⁡x and cos⁡x\sin x \ \text{and} \ \cos x are even
  • Convert everything into sin⁡(...x)\sin\left(...x\right) or cos⁡(...x)\cos\left(...x\right) using double and half angle formulas:
  • sin⁡2x=1−cos⁡2x2\displaystyle \sin^2x=\frac{1-\cos2x}{2}
  • cos⁡2x=1+cos⁡2x2\displaystyle \cos^2x=\frac{1+\cos2x}{2}
  • sin⁡xcos⁡x=sin⁡2x2\displaystyle \sin x\cos x=\frac{\sin2x}{2}
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Case 2: ∫tan⁡mxsec⁡nx  ⁣dx{\displaystyle\int}\tan^m x\sec^nx\de{x}

  • Power of sec⁡x\sec x is even
  • Keep a copy of sec⁡2x\sec^2x and convert the rest to tan⁡x\tan x using sec⁡2x=1+tan⁡2x\sec^2x=1+\tan^2x
  • Formula: ∫tan⁡mx(1+tan⁡2x)ksec⁡2x  ⁣dx{\displaystyle \int}\tan^mx(1+\tan^2x)^k\sec^2 x\de{x}
  • Do a U-substitution with u=tan⁡xu=\tan x
  • Power of tan⁡x\tan x is odd
  • Keep a copy of sec⁡x tan⁡x\sec x\ \tan x and convert the rest to sec⁡x\sec x using tan⁡2x=sec⁡2x−1\tan^2x=\sec^2x-1
  • Formula: ∫sec⁡n−1x(sec⁡2x−1)ksec⁡xtan⁡x  ⁣dx{\displaystyle \int}\sec^{n-1}x(\sec^2x-1)^k\sec x\tan x\de{x}
  • Do a U-substitution with u=sec⁡xu=\sec x

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Case 3: ∫cot⁡mxcsc⁡nx  ⁣dx{\displaystyle\int}\cot^m x\csc^nx\de{x}

  • Power of csc⁡x\csc x is even
  • Keep a copy of csc⁡2x\csc^2x and convert the rest to cot⁡x\cot x using csc⁡2x=1+cot⁡2x\csc^2x=1+\cot^2x
  • Formula: ∫cot⁡mx(1+cot⁡2x)kcsc⁡2x  ⁣dx{\displaystyle \int}\cot^mx(1+\cot^2x)^k\csc^2x\de{x}
  • Do a U-substitution with u=cot⁡xu=\cot x
  • Power of cot⁡x\cot x is odd
  • Keep a copy of csc⁡x cot⁡x\csc x\ \cot x and convert the rest to csc⁡x\csc x using cot⁡2x=csc⁡2x−1\cot^2x=\csc^2x-1
  • Formula: ∫csc⁡n−1x(csc⁡2x−1)kcsc⁡xcot⁡x  ⁣dx{\displaystyle \int}\csc^{n-1}x(\csc^2x-1)^k\csc x\cot x\de{x}
  • Do a U-substitution with u=csc⁡xu=\csc x
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Case 4: ∫sin⁡(mx) cos⁡(nx)   ⁣dx ,∫sin⁡(mx) sin⁡(nx)   ⁣dx,∫cos⁡(mx) cos⁡(nx)   ⁣dx{\displaystyle\int}\sin(mx)\ \cos (nx)\ \de{x} \ ,{\displaystyle\int}\sin(mx)\ \sin(nx)\ \de{x}, {\displaystyle\int}\cos(mx)\ \cos(nx)\ \de{x}

Use the identities
  • sin⁡A cos⁡B=12[sin⁡(A−B)+sin⁡(A+B)]\displaystyle \sin A\ \cos B=\frac{1}{2}\left[\sin\left(A-B\right)+\sin\left(A+B\right)\right]
  • sin⁡A sin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\displaystyle \sin A\ \sin B=\frac{1}{2}\left[\cos\left(A-B\right)-\cos\left(A+B\right)\right]
  • cos⁡A cos⁡B=12[cos⁡(A−B)+cos⁡(A+B)]\displaystyle \cos A\ \cos B=\frac{1}{2}\left[\cos\left(A-B\right)+\cos\left(A+B\right)\right]
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Example: Trig Integrals (Case 1)

Evaluate the following indefinite integral

∫sin⁡4xcos⁡3x   ⁣dx{\displaystyle\int}\sin^4x\cos^3x\space\de{x}

Keep one copy of cos and convert the rest to sin
=∫sin⁡4x(cos⁡2x) (cos⁡x)   ⁣dx={\displaystyle\int}\sin^4x(\cos^2x)\space(\cos x)\space\de{x}
=∫sin⁡4x(1−sin⁡2x)(cos⁡x)   ⁣dx={\displaystyle\int} \sin^4x(1-\sin^2x)(\cos x) \space\de{x}

We need to do a U-substitution:

Let u=sin⁡xu=\sin x, then   ⁣du=cos⁡x   ⁣dx\de{u}=\cos x \space\de{x}, and dx=1cos⁡xdu\displaystyle dx=\frac{1}{\cos x}du
The new integral becomes:
=∫u4(1−u2)(cos⁡x)(1cos⁡x  ⁣du)=\displaystyle\int u^4(1-u^2)(\cos x)\left(\frac{1}{\cos x}\de{u}\right)

=∫u4(1−u2) du\displaystyle =\int_{ }^{ }u^4\left(1-u^2\right)\ du

=∫(u4−u6)du\displaystyle =\int_{ }^{ }\left(u^4-u^6\right)du

=u55−u77+C\displaystyle =\frac{u^5}{5}-\frac{u^7}{7}+C

=sin⁡5x5−sin⁡7x7+C\displaystyle =\frac{\sin^5x}{5}-\frac{\sin^7x}{7}+C
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Example: Trig Integrals (Case 1)

Evaluate the following indefinite integral

∫sin⁡2xcos⁡2x   ⁣dx{\displaystyle\int} \sin^2x\cos^2x\space\de{x}
Let's convert everything into cos x using the half-angle formulas:
=∫[1−cos⁡(2x)2][1+cos⁡(2x)2]  ⁣dx\displaystyle ={\displaystyle\int}\left[\frac{1-\cos(2x)}{2}\right]\left[\frac{1+\cos(2x)}{2}\right]\de{x}

=∫1−cos⁡2(2x)22  ⁣dx\displaystyle ={\displaystyle\int}\frac{1-\cos^2(2x)}{2^2}\de{x}

=∫(14−cos⁡2(2x)4) dx\displaystyle ={\displaystyle \int}\left(\frac{1}{4}-\frac{\cos^2\left(2x\right)}{4}\right)\ dx

=∫14  ⁣dx − 14∫cos⁡2(2x)  ⁣dx\displaystyle ={\displaystyle\int}\frac{1}{4}\de{x}\space-\space\frac{1}{4}{\displaystyle\int}\cos^2(2x)\de{x}

Using the half-angle formula again for cos:

=∫14  ⁣dx − 14∫1+cos⁡(4x)2  ⁣dx\displaystyle ={\displaystyle\int}\frac{1}{4}\de{x}\space-\space\frac{1}{4}{\displaystyle\int}\frac{1+\cos(4x)}{2}\de{x}

=∫14  ⁣dx − 18∫(1+cos⁡(4x))  ⁣dx\displaystyle ={\displaystyle\int}\frac{1}{4}\de{x}\space - \space\frac{1}{8}{\displaystyle\int}(1+\cos(4x))\de{x}

=14x − 18x − sin⁡(4x)32+C\displaystyle =\frac{1}{4}x \space- \space\frac{1}{8}x \space - \space\frac{\sin(4x)}{32}+C

=x8 − sin⁡(4x)32+C\displaystyle =\frac{x}{8} \space - \space \frac{\sin(4x)}{32}+C
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Example: Trig Integrals (Case 2)

Evaluate the following indefinite integral

∫sec⁡8xtan⁡6x   ⁣dx{\displaystyle\int}\sec^8x\tan^6x\space\de{x}
Keep a copy of sec⁡2x\sec^2x and convert the rest to tan⁡x\tan x:

=∫(sec⁡2x)(sec⁡6x)tan⁡6x dx\displaystyle =\int_{ }^{ }\left(\sec^2x\right)\left(\sec^6x\right)\tan^6x\ dx
=∫(sec⁡2x)(sec⁡2x)3tan⁡6x dx\displaystyle =\int_{ }^{ }\left(\sec^2x\right)\left(\sec^2x\right)^3\tan^6x\ dx
=∫(sec⁡2x)(1+tan⁡2x)3tan⁡6x dx\displaystyle =\int_{ }^{ }\left(\sec^2x\right)\left(1+\tan^2x\right)^3\tan^6x\ dx

We need to do a U-substitution:

Let u=tan⁡xu=\tan x, then   ⁣du=sec⁡2x  ⁣dx\de{u}=\sec^2x\de{x} and dx=1sec⁡2xdu\displaystyle dx=\frac{1}{\sec^2x}du
The new integral becomes:
=∫(sec⁡2x)(1+u2)3(u6)(1sec⁡2xdu)\displaystyle =\int_{ }^{ }\left(\sec^2x\right)\left(1+u^2\right)^3\left(u^6\right)\left(\frac{1}{\sec^2x}du\right)
=∫(1+u2)3 (u6) du\displaystyle =\int_{ }^{ }\left(1+u^2\right)^{3\ }\left(u^6\right)\ du
=∫(u6+3u4+3u2+1) u6 du\displaystyle =\int_{ }^{ }\left(u^6+3u^4+3u^2+1\right)\ u^6\ du
=∫(u12+3u10+3u8+u6) du\displaystyle =\int_{ }^{ }\left(u^{12}+3u^{10}+3u^8+u^6\right)\ du
=u1313+3u1111+3u99+u77+C\displaystyle =\frac{u^{13}}{13}+\frac{3u^{11}}{11}+\frac{3u^9}{9}+\frac{u^7}{7}+C
=tan⁡13x13+3tan⁡11x11+tan⁡9x3+tan⁡7x7+C\displaystyle =\frac{\tan^{13}x}{13}+\frac{3\tan^{11}x}{11}+\frac{\tan^9x}{3}+\frac{\tan^7x}{7}+C

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Example: Trig Integrals (Case 4)

Evaluate the following indefinite integral

∫sin⁡(3x)sin⁡(2x)   ⁣dx{\displaystyle \int} \sin(3x)\sin(2x)\ \de{x}

Using the appropriate identity:

∫sin⁡(3x)sin⁡(2x)   ⁣dx{\displaystyle \int} \sin(3x)\sin(2x)\ \de{x}
=∫12[cos⁡(3x−2x)−cos⁡(3x+2x)]  ⁣dx\displaystyle ={\displaystyle \int} \frac{1}{2}[\cos(3x-2x)-\cos(3x+2x)]\de{x}

=12∫[cos⁡x−cos⁡5x]  ⁣dx\displaystyle =\frac{1}{2}{\displaystyle \int} [\cos x-\cos 5x]\de{x}

=12[sin⁡x−sin⁡x5]+C\displaystyle =\frac{1}{2}\left[\sin x-\frac{\sin x}{5}\right]+C

=sin⁡x2−sin⁡5x10+C\displaystyle =\frac{\sin x}{2}-\frac{\sin5x}{10}+C
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Example: Trig Integrals

Evaluate the following indefinite integral

∫sec⁡x dx\displaystyle \int_{ }^{ }\sec x\ dx

=∫(sec⁡x ⋅ sec⁡x+tan⁡xsec⁡x+tan⁡x)dx\displaystyle ={\displaystyle \int}\left(\sec x\ \cdot\ \frac{\sec x+\tan x}{\sec x+\tan x}\right)dx

=∫(sec⁡2x+sec⁡x tan⁡xsec⁡x+tan⁡x)dx\displaystyle ={\displaystyle \int}\left(\frac{\sec^2x+\sec x\ \tan x}{\sec x+\tan x}\right)dx

We'll do a U-substitution:

Let u=sec⁡x+tan⁡xu=\sec x+\tan x, then du=(sec⁡x tan⁡x+sec⁡2x)dxdu=\left(\sec x\ \tan x+\sec^2x\right)dx, and dx=1sec⁡x tan⁡x+sec⁡2xdu\displaystyle dx=\frac{1}{\sec x\ \tan x+\sec^2x}du
The new integral becomes:

=∫(sec⁡2x+sec⁡x tan⁡xu)(1sec⁡x tan⁡x+sec⁡2xdu)\displaystyle ={\displaystyle \int}\left(\frac{\sec^2x+\sec x\ \tan x}{u}\right)\left( \frac{1}{\sec x\space \tan x+\sec^2x}du\right)

=∫1u du\displaystyle ={\displaystyle \int}\frac{1}{u}\space du

=ln⁡∣u∣+C\displaystyle =\ln\left|u\right|+C

=ln⁡∣sec⁡x+tan⁡x∣+C=\ln\left|\sec x+\tan x\right|+C

Practice: Trig Integrals

Evaluate the following indefinite integral

∫cos⁡4(x)  ⁣dx{\displaystyle \int}\cos^4(x)\de{x}

Practice: Trig Integrals

Evaluate the following indefinite integral

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

Practice: Trig Integrals

Evaluate the following indefinite integral

∫sec⁡6x  ⁣dx{\displaystyle\int}\sec^6x\de{x}

Practice: Trig Integrals

Evaluate the following indefinite integral

∫sin⁡(15x)cos⁡(4x)  ⁣dx{\displaystyle\int}\sin(15x)\cos(4x)\de{x}