Ask Questions, Get Answers

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

Evaluate the definite integrals\[\int\limits_\frac{\pi}{2}^\pi e^x\bigg(\frac{1-\sin x}{1-\cos x}\bigg)dx\]

Can you answer this question?

1 Answer

0 votes
  • (i)$\int e^x[f(x)+f'(x)]dx=e^x f(x)$
  • (ii)$\sin^2x+\cos^2x=1$
  • (iii)$\sin 2x=2 \sin x\cos x$
  • (iv)$2 \sin ^2 x/2=1-\cos x$
Given $I=\int\limits_\frac{\pi}{2}^\pi e^x\bigg[\frac{1-\sin x}{1-\cos x}\bigg]dx$
We can write $\sin x\;as\;2\sin x/2 \cos x/2\; and\; t-\cos x \;as\;2\sin^2x/2$
Therefore $ I=\int\limits_\frac{\pi}{2}^\pi e^x\bigg[\frac{1-2\sin x/2-\cos x/2}{2 \sin^2 x/2}\bigg]dx$
On spliting the terms we get,
$\int\limits_\frac{\pi}{2}^\pi e^x\bigg[\frac{1}{2\sin^2 x/2}-\cot x/2\bigg]dx$
But $\frac{1}{\sin ^2 x/2}=cosec ^2 x/2$
Hence $I=\int\limits_\frac{\pi}{2}^\pi e^x\bigg[\frac{cosec ^2 x/2}{2}-\cot x/2\bigg]dx$
Let $ -\cot x/2=f(x)\qquad => f'(x)=-(-\frac{1}{2} cosec^2 x/2)$
Therefore $=\int\limits_\frac{\pi}{2}^\pi e^x\bigg[f(x)+f'(x)]dx=[e^x(f(x)\bigg]^\pi_{\pi/2}$
Therefore $I=e^x[\cot x/2]^\pi_{\pi/2}$
On applying the limits we get,
$I=e^\pi \cos{\pi/2}-e^{\pi/2} \cot \pi/4$
But $ \cot \pi/2=0\;and\;\cot \pi/4=1$
Therefore $I=e^\pi \times 0-e^{\pi/2} \times 1$
Hence $\int\limits_\frac{\pi}{2}^\pi e^x\bigg[\frac{1-\sin x}{1-\cos x}\bigg]dx=e^{\pi/2}$



answered Feb 18, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App