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

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Toolbox:
  • (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 $I=e^\pi \times 0-e^{\pi/2} \times 1$
 
=$e^{\pi/2}$
 
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
 

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