Ask Questions, Get Answers

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

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

0 votes
  • (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.
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
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App