Integration TechniquesTopic #24 of 32

Trigonometric Integrals

Techniques for integrating powers and products of trig functions.

Powers of Sine and Cosine

∫sin⁡n(x)cos⁡m(x) dx\int \sin^n(x) \cos^m(x) \, dx

Case 1: n is odd (n=2k+1n = 2k + 1)

Save one sin⁡(x)\sin(x) for dudu, convert rest using sin⁡2(x)=1−cos⁡2(x)\sin^2(x) = 1 - \cos^2(x):

∫sin⁡2k+1(x)cos⁡m(x) dx=∫(1−cos⁡2x)kcos⁡m(x)sin⁡(x) dx\int \sin^{2k+1}(x) \cos^m(x) \, dx = \int (1-\cos^2 x)^k \cos^m(x) \sin(x) \, dx

Let u=cos⁡(x)u = \cos(x), du=−sin⁡(x) dxdu = -\sin(x) \, dx

Case 2: m is odd (m=2k+1m = 2k + 1)

Save one cos⁡(x)\cos(x) for dudu, convert rest using cos⁡2(x)=1−sin⁡2(x)\cos^2(x) = 1 - \sin^2(x):

∫sin⁡n(x)cos⁡2k+1(x) dx=∫sin⁡n(x)(1−sin⁡2x)kcos⁡(x) dx\int \sin^n(x) \cos^{2k+1}(x) \, dx = \int \sin^n(x) (1-\sin^2 x)^k \cos(x) \, dx

Let u=sin⁡(x)u = \sin(x), du=cos⁡(x) dxdu = \cos(x) \, dx

Case 3: Both n and m are even

Use half-angle formulas:

sin⁡2(x)=1−cos⁡(2x)2\sin^2(x) = \frac{1 - \cos(2x)}{2}

cos⁡2(x)=1+cos⁡(2x)2\cos^2(x) = \frac{1 + \cos(2x)}{2}

sin⁡(x)cos⁡(x)=sin⁡(2x)2\sin(x)\cos(x) = \frac{\sin(2x)}{2}

Powers of Tangent and Secant

∫tan⁡n(x)sec⁡m(x) dx\int \tan^n(x) \sec^m(x) \, dx

Case 1: m is even (m=2km = 2k)

Save sec⁡2(x)\sec^2(x) for dudu, convert rest using sec⁡2(x)=1+tan⁡2(x)\sec^2(x) = 1 + \tan^2(x):

∫tan⁡n(x)sec⁡2k(x) dx=∫tan⁡n(x)(1+tan⁡2x)k−1sec⁡2(x) dx\int \tan^n(x) \sec^{2k}(x) \, dx = \int \tan^n(x) (1+\tan^2 x)^{k-1} \sec^2(x) \, dx

Let u=tan⁡(x)u = \tan(x), du=sec⁡2(x) dxdu = \sec^2(x) \, dx

Case 2: n is odd (n=2k+1n = 2k + 1)

Save sec⁡(x)tan⁡(x)\sec(x)\tan(x) for dudu, convert rest using tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1:

∫tan⁡2k+1(x)sec⁡m(x) dx=∫(sec⁡2x−1)ksec⁡m−1(x)sec⁡(x)tan⁡(x) dx\int \tan^{2k+1}(x) \sec^m(x) \, dx = \int (\sec^2 x-1)^k \sec^{m-1}(x) \sec(x)\tan(x) \, dx

Let u=sec⁡(x)u = \sec(x), du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x) \, dx

Important Reduction Formulas

∫sin⁡n(x) dx=−sin⁡n−1(x)cos⁡(x)n+n−1n∫sin⁡n−2(x) dx\int \sin^n(x) \, dx = -\frac{\sin^{n-1}(x)\cos(x)}{n} + \frac{n-1}{n} \int \sin^{n-2}(x) \, dx

∫cos⁡n(x) dx=cos⁡n−1(x)sin⁡(x)n+n−1n∫cos⁡n−2(x) dx\int \cos^n(x) \, dx = \frac{\cos^{n-1}(x)\sin(x)}{n} + \frac{n-1}{n} \int \cos^{n-2}(x) \, dx

∫tan⁡n(x) dx=tan⁡n−1(x)n−1−∫tan⁡n−2(x) dx\int \tan^n(x) \, dx = \frac{\tan^{n-1}(x)}{n-1} - \int \tan^{n-2}(x) \, dx

∫sec⁡n(x) dx=sec⁡n−2(x)tan⁡(x)n−1+n−2n−1∫sec⁡n−2(x) dx\int \sec^n(x) \, dx = \frac{\sec^{n-2}(x)\tan(x)}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}(x) \, dx

Examples

Example 1: Odd power of sine

∫sin⁡3(x) dx=∫sin⁡2(x)⋅sin⁡(x) dx=∫(1−cos⁡2x)sin⁡(x) dx\int \sin^3(x) \, dx = \int \sin^2(x) \cdot \sin(x) \, dx = \int (1-\cos^2 x)\sin(x) \, dx

Let u=cos⁡(x)u = \cos(x), du=−sin⁡(x) dxdu = -\sin(x) \, dx

=−∫(1−u2) du=−u+u33+C=−cos⁡(x)+cos⁡3(x)3+C= -\int (1-u^2) \, du = -u + \frac{u^3}{3} + C = -\cos(x) + \frac{\cos^3(x)}{3} + C

Example 2: Both powers even

∫sin⁡2(x)cos⁡2(x) dx=∫[sin⁡(x)cos⁡(x)]2 dx\int \sin^2(x)\cos^2(x) \, dx = \int [\sin(x)\cos(x)]^2 \, dx

=∫[sin⁡(2x)2]2dx=14∫sin⁡2(2x) dx= \int \left[\frac{\sin(2x)}{2}\right]^2 dx = \frac{1}{4} \int \sin^2(2x) \, dx

=14∫1−cos⁡(4x)2 dx=18[x−sin⁡(4x)4]+C= \frac{1}{4} \int \frac{1-\cos(4x)}{2} \, dx = \frac{1}{8}\left[x - \frac{\sin(4x)}{4}\right] + C

Example 3: Even power of secant

∫sec⁡4(x) dx=∫sec⁡2(x)⋅sec⁡2(x) dx=∫(1+tan⁡2x)sec⁡2(x) dx\int \sec^4(x) \, dx = \int \sec^2(x) \cdot \sec^2(x) \, dx = \int (1+\tan^2 x)\sec^2(x) \, dx

Let u=tan⁡(x)u = \tan(x), du=sec⁡2(x) dxdu = \sec^2(x) \, dx

=∫(1+u2) du=u+u33+C=tan⁡(x)+tan⁡3(x)3+C= \int (1+u^2) \, du = u + \frac{u^3}{3} + C = \tan(x) + \frac{\tan^3(x)}{3} + C

Example 4: Odd power of tangent

∫tan⁡3(x) dx=∫tan⁡(x)⋅tan⁡2(x) dx=∫tan⁡(x)(sec⁡2x−1) dx\int \tan^3(x) \, dx = \int \tan(x) \cdot \tan^2(x) \, dx = \int \tan(x)(\sec^2 x - 1) \, dx

=∫tan⁡(x)sec⁡2(x) dx−∫tan⁡(x) dx= \int \tan(x)\sec^2(x) \, dx - \int \tan(x) \, dx

For first integral, let u=tan⁡(x)u = \tan(x):

=tan⁡2(x)2−ln⁡∣sec⁡(x)∣+C= \frac{\tan^2(x)}{2} - \ln|\sec(x)| + C

Example 5: Mixed tan and sec

∫tan⁡3(x)sec⁡3(x) dx\int \tan^3(x)\sec^3(x) \, dx

Save sec⁡(x)tan⁡(x)\sec(x)\tan(x), use tan⁡2=sec⁡2−1\tan^2 = \sec^2 - 1:

=∫tan⁡2(x)sec⁡2(x)⋅sec⁡(x)tan⁡(x) dx= \int \tan^2(x)\sec^2(x) \cdot \sec(x)\tan(x) \, dx

=∫(sec⁡2x−1)sec⁡2(x)⋅sec⁡(x)tan⁡(x) dx= \int (\sec^2 x - 1)\sec^2(x) \cdot \sec(x)\tan(x) \, dx

Let u=sec⁡(x)u = \sec(x), du=sec⁡(x)tan⁡(x) dxdu = \sec(x)\tan(x) \, dx

=∫(u2−1)u2 du=∫(u4−u2) du= \int (u^2 - 1)u^2 \, du = \int (u^4 - u^2) \, du

=u55−u33+C=sec⁡5(x)5−sec⁡3(x)3+C= \frac{u^5}{5} - \frac{u^3}{3} + C = \frac{\sec^5(x)}{5} - \frac{\sec^3(x)}{3} + C