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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\sin^3(2x+1)\]

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


answered Jan 30, 2013 by sreemathi.v
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