Browse Questions

Find the integrals of the functions$\sin^3(2x+1)$

Toolbox:
• (i)$\sin^2x=\frac{1-\cos 2x}{2}$
• (ii)$\sin^3x=\frac{3\sin x-\sin 3x}{4}$
• (iii)Method of substituting
• If $f(x)=t.\Rightarrow f'(x)dx=dt.$
• $Thus\; I=\int f(x)dx=\int t dt.$
Given $I=\int \sin^3(2x+1)dx.$

using the information in the toolbox we can write

$\sin^3x=\frac{(3\sin x-\sin 3x)}{4}$

$I=\int\frac{3\sin(2x+1)-\sin 3(2x+1)}{4}dx.$

On separating we get,

$I=\frac{3}{4}\int \sin(2x+1)dx-\frac{1}{4}\int\sin 3(2x+1)dx$

Let 2x+1=t.

On differentiating we get,

2dx=dt$\Rightarrow dx=\frac{dt}{2}$

On substituting t and dt we get,

Hence $I=\frac{3}{4}\int\sin t.\frac{dt}{2}-\int\frac{1}{4}\sin 3t.\frac{dt}{2}$

$\;\;\;=\frac{3}{8}\int\sin t-\frac{1}{8}\int\sin 3t.dt$

On integrating we get,

$\;\;\;=\frac{-3}{8}\cos t-[\frac{1}{8}(\frac{-1}{3}\cos 3t)]+c.$

On multiplying we get,

$\;\;\;=\frac{-3}{8}\cos t+\frac{1}{24}\cos3 t+c.$

Taking $\frac{3}{8}$ as a common factor we get,

$\;\;\;=\frac{3}{8}[-\cos t+\frac{1}{3}\cos 3t]+c.$

Substituting for t we get,

$\;\;\;=\frac{3}{8}[\frac{1}{3}\cos 3(2x+1)-\cos(2x+1)]+c$.