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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[e^x\bigg(\frac{1+\sin x}{1+\cos x}\bigg)\]

$\begin{array}{1 1} e^x \tan \large\frac{x}{2} + c \\ e^x \sin \large\frac{x}{2} + c \\e^x \cos \large\frac{x}{2} + c \\ e^x \cot \large\frac{x}{2} + c \end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • (i)$\int e^x\{f(x)+f'(x)\}dx=e^x(f(x))+c.$
  • (ii)$1+\sin x=\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin x\cos x=(\sin\frac{x}{2}+\cos \frac{x}{2})^2.$.$
  • (iii)$1+\cos x=2\cos^2\frac{x}{2}.$
Given $I=\int e^x\begin{bmatrix}\frac{1+\sin x}{1+\cos x}\end{bmatrix}dx.$
 
Using the information in the toolbox,we can write I as,
 
$I=\int e^x\frac{(\sin x/2+\cos x/2)^2}{2\cos^2x/2}dx.$
 
$\;\;\;=\frac{1}{2}\int e^x\begin{bmatrix}\frac{\sin x/2+\cos x/2}{\cos x/2}\end{bmatrix}dx.$
 
Now separating the terms,
 
$I=\frac{1}{2}\int e^x(\tan x/2+1)^2.$
 
On expanding,
 
$I=\frac{1}{2}\int e^x(\tan^2x/2+1+2\tan x/2)dx.$
 
But $1+\tan ^2x/2=\sec^2x/2.$
 
$I=\int e^x.\frac{1}{2}(\sec^2x/2)+(\tan x/2)dx.$
 
Let f(x)=tan x/2.
 
f'(x)=$\frac{1}{2}\sec^2x/2.$
 
Hence $I=\int e^x\{f(x)+f'(x)\}dx=e^xf(x)+c.$
 
$\qquad\;\;\;\;=e^x\tan x/2+c.$

 

answered Feb 11, 2013 by sreemathi.v
 
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