logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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

Can you answer this question?
 
 

1 Answer

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

 

answered Jan 30, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...