Integrating Trigonometric Functions by Double Angle Formula

Integrating Trigonometric Functions by Double Angle Formula

Integrating Trigonometric Functions can be done by Double Angle Formula reducing the power of trigonometric functions.

\begin{aligned} \displaystyle \cos{2A} &= 2\cos^2{A}-1 \\ &= 1-2\sin^2{A} \\ &= \cos^2{A}-\sin^2{A} \end{aligned}

Practice Questions

Question 1

Find $\displaystyle \int{\cos^2{x}}dx$.

\begin{aligned} \displaystyle 2\cos^2{x}-1 &= \cos{2x} \\ 2\cos^2{x} &= \cos{2x} + 1 \\ \cos^2{x} &= \frac{1}{2}(\cos{2x}+1) \\ \int{\cos^2{x}}dx &= \frac{1}{2} \int{(\cos{2x}+1)}dx \\ &= \frac{1}{2}\Big(\frac{1}{2}\sin{2x}+x\Big) + C \\ &= \frac{1}{4} \sin{2x} + \frac{1}{2}x + C \end{aligned}

Question 2

Find $\displaystyle \int{\sin^2{x}}dx$.

\begin{aligned} \displaystyle \cos{2x} &= 1-2\sin^2{x} \\ 2\sin^2{x} &= 1-\cos{2x} \\ \sin^2{x} &= \frac{1}{2}(1-\cos{2x}) \\ \int{\sin^2{x}}dx &= \frac{1}{2} \int{(1-\cos{2x})}dx \\ &= \frac{1}{2}\Big(x-\frac{1}{2}\sin{2x}\Big) + C \\ &= \frac{1}{2}x-\frac{1}{4}\sin{2x} + C \end{aligned}

Question 3

Find $\displaystyle \int{\sin^4{x}}dx$.

\begin{aligned} \displaystyle \sin^4{x} &= (\sin^2{x})^2 \\ &= \frac{1}{4}(1-\cos{2x})^2 \\ &= \frac{1}{4}(1-2\cos{2x}+\cos^2{2x}) \\ &= \frac{1}{4}\Big[1-2\cos{2x}+\frac{1}{2}(1+\cos{4x})\Big] \\ &= \frac{1}{4}\Big[\frac{3}{2}-2\cos{2x}+\frac{1}{2}\cos{4x}\Big] \\ &= \frac{3}{8}-\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x} \\ \int{\sin^4{x}}dx &= \int{\Big(\frac{3}{8}-\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}\Big)}dx \\ &= \frac{3}{8}x-\frac{1}{4}\sin{2x} + \frac{1}{32}\sin{4x} + C \end{aligned}

Question 4

Find $\displaystyle \int{(\cos^2x-\sin^2x)}dx$.

\begin{aligned} \displaystyle \int{(\cos^2x-\sin^2x)}dx &= \int{\cos{2x}}dx \\ &= \frac{1}{2}\sin{2x} + C \end{aligned}