6CUsingTrigIdentities

Integration of Trig Identities

You have already seen the integration of basic trig functions:

math \\ . \qquad \displaystyle{ \int{ \sin kx }\; dx} = -\frac{1}{k} \cos kx + c \\. \\ . \qquad \displaystyle{ \int{ \cos kx }\; dx} = \frac{1}{k} \sin kx + c \\. \\ . \qquad \displaystyle{ \int{ \sec^2 kx }\: dx} = \frac{1}{k} \tan kx + c math

To integrate many trig indentities we use the double angle formulas backwards to express the identities in terms of one of the basic trig functions. . math . \qquad \sin(2A) = 2\sin(A) \cos(A) math . math \\ . \qquad \cos(2A) = \cos^2(A) - \sin^2(A) \\ .\\ . \qquad \cos(2A) = 2\cos^2(A) - 1 \\. \\ . \qquad \cos(2A) = 1 - 2\sin^2(A) math . math . \qquad \tan(2A) = \dfrac {2\tan(A)}{1-\tan^2(A)} math

Example 1

math . \qquad \displaystyle{ \int{ \sin^2 (3x) }\;dx} math

math \\ . \qquad \text {Use } \; \cos(2A) = 1 - 2\sin^2(A) \\. \\ . \qquad \Rightarrow \quad \sin^2(A) = \dfrac{1}{2} \big( 1 - \cos(2A) \big) math

Hence

math \\ . \qquad \displaystyle{ \int{ \sin^2 (3x) }\; dx} \\. \\ . \qquad = \displaystyle{ \int{ \dfrac{1}{2} \big( 1 - \cos(6x) \big) }\;dx} math . math \\ . \qquad = \dfrac{1}{2} \displaystyle{ \int{ 1 - \cos(6x) }\;dx} \\. \\ . \qquad = \dfrac{1}{2} \left( x - \dfrac{1}{6} \sin (6x) \right) + c \\. \\ . \qquad = \dfrac{x}{2} - \dfrac{1}{12} \sin (6x) + c math

Example 2

math . \qquad \displaystyle{ \int{ 5 \sin (3x) \cos (3x) }\; dx} math

math \\ . \qquad \text {Use } \; \sin(2A) = 2\sin(A)\cos(A) \\. \\ . \qquad \Rightarrow \quad \sin(A)\cos(A) = \dfrac{1}{2} \sin(2A) math

Hence

math \\ . \qquad \displaystyle{ \int{ 5 \sin (3x) \cos (3x) }\;dx} \\. \\ . \qquad = \displaystyle{ \int{ 5 \times \dfrac{1}{2} \sin(6x) }\;dx} math . math \\ . \qquad = \dfrac{5}{2} \displaystyle{ \int{ \sin(6x) }\; dx} \\. \\ . \qquad = \dfrac{5}{2} \left( - \dfrac{1}{6} \cos (6x) \right) + c \\. \\ . \qquad = - \dfrac{5}{12} \cos (6x) + c math

Even Powers of Sin or Cos

Use the double angle formula (more than once if necessary)

Example

math . \qquad \displaystyle{ \int{ \sin^4 (2x) }\; dx} math

math . \qquad \text {Use } \; \cos(2A) = 1 - 2 \sin^2 (A) math

math \\ . \qquad \displaystyle{ \int{ \sin^4 (2x) } \; dx} \\. \\ . \qquad = \displaystyle{ \int{ \Big( \dfrac{1}{2} \big( 1 - \cos (4x) \big) \Big)^2 }\; dx} \\. \\ . \qquad = \displaystyle{ \int{ \dfrac{1}{4} \big( 1-\cos (4x) \big)^2 }\; dx} math . math . \qquad = \dfrac{1}{4} \displaystyle{ \int{ 1 - 2\cos (4x) +\cos^2 (4x) }\;dx } math . math \text {Now use } \; \cos(2A) = 2\cos^2(A) - 1 \; \text { on the last term} math . math \\ . \qquad = \dfrac{1}{4} \displaystyle{ \int{ 1 - 2\cos (4x) + \dfrac{1}{2} \big( 1 + \cos (8x) \big) }\; dx} \\. \\ . \qquad = \dfrac{1}{4} \displaystyle{ \int{ 1 - 2\cos (4x) + \dfrac{1}{2} + \dfrac{1}{2} \cos (8x) }\; dx} math . math \\ . \qquad = \dfrac{1}{4} \displaystyle{ \int{ \dfrac{3}{2} - 2\cos (4x) + \dfrac{1}{2} \cos (8x) }\;dx} \\. \\ . \qquad = \dfrac{1}{4} \left( \dfrac{3x}{2} - \dfrac{2}{4}\sin (4x) + \dfrac{1}{16} \sin (8x) \right) + c \\ .\\ . \qquad = \dfrac{3x}{8} - \dfrac{1}{8}\sin (4x) + \dfrac{1}{64} \sin (8x) + c math

**See also:** www.mathsonline.com.au Y12Extension --> Calculus --> Methods of Integration --> Lesson 3

Odd Powers of Sin or Cos
 * 1) Separate a single instance of the trig function (sin or cos)
 * 2) Use cos 2 (A) + sin 2 (A) = 1 on the remainder to swap the remainder to the other function
 * 3) Substitute using u = the other trig function (sin or cos) from that chosen in step 1

Example

math . \qquad \displaystyle{ \int{ \sin^3 (x) }\;dx} math

1. Separate a single instance of the sin function

math \\ . \qquad \displaystyle{ \int{ \sin^3 (x) } \; dx} \\. \\ . \qquad = \displaystyle{ \int{ \big( \sin (x) \big) \big( \sin^2 (x) \big) }\;dx} math

2. Use sin 2 (A) = 1 - cos 2 (A) on the remainder

math . \qquad = \displaystyle{ \int{ \big( \sin (x) \big) \big( 1 - \cos^2 (x) \big) }\; dx} math

3. Let u = cos x (step 1 was sin x) and use the substitution method.

math \\ . \qquad \text{Use } \; u = \cos (x) \\. \\ . \qquad \Rightarrow \; \dfrac {du}{dx} = - \sin (x) \\. \\ . \qquad \Rightarrow \; dx = - \dfrac {du}{\sin (x)} math

Hence

math \\ . \qquad \displaystyle{ \int{ \sin^3 (x) }\;dx} \\. \\ . \qquad = \displaystyle{ \int{ \big( \sin (x) \big) \big( 1 - u^2 \big) } \; \left( - \dfrac{du}{\sin (x)} \right) } math . math . \qquad = \displaystyle{ \int{ - \big( 1 - u^2 \big) }\;du } math . math . \qquad = \displaystyle{ \int{ u^2 - 1 }\;du} math . math \\ . \qquad = \dfrac {u^3}{3} - u + c \\. \\ . \qquad = \dfrac {\cos^3(x)}{3} - \cos (x) + c math

More Examples .