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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $\int_0^{\large\frac{\pi}{2}} \large\frac{x+sinx}{1+cosx}$$dx $

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1 Answer

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Toolbox:
  • $\int udv=uv-\int vdu$
  • $\tan\large\frac{\pi}{4}$$=1$
  • $\sin x=2\sin \Large\frac{x}{2}$$\cos\Large\frac{x}{2}$
  • $1+\cos x=2\cos^2\Large\frac{x}{2}$
Step 1:
$I=\int_0^{\large\frac{\pi}{2}}\large\frac{x+\sin x}{1+\cos x}$$dx$
But $1+\cos x=2\cos^2\Large\frac{x}{2}$
$\sin x=2\sin \Large\frac{x}{2}$$\cos\Large\frac{x}{2}$
$I=\large\frac{1}{2}\int_0^{\Large\frac{\pi}{2}}$$x\sec^2\Large\frac{x}{2}$$dx+\int_0^{\Large\frac{\pi}{2}}\tan\large\frac{x}{2}$$dx$
Let $I=I_1+I_2$
Step 2:
On integrating we get,
Consider $I_1=\int x\sec^2\Large\frac{x}{2}$$dx$
Let $u=x\Rightarrow du=dx$
$\sec^2\Large\frac{x}{2}$$dx=dv$
$2\tan \Large\frac{x}{2}$$=v$
$\int udv=uv-\int vdu$
On substituting for $u,v,du$ and $dv$ we get,
$(x)(2\tan\Large\frac{x}{2})$$-\int 2\tan \Large\frac{x}{2}$$dx$
$\therefore I=\large\frac{1}{2}$$\big[2x\tan\Large\frac{x}{2}\big]_0^{\Large\frac{\pi}{2}}-\frac{1}{2}$$\times 2\int_0^{\Large\frac{\pi}{2}}\tan x/2dx+\int_0^{\Large\frac{\pi}{2}}\tan x/2dx$
$\big[x\tan\Large\frac{x}{2}\big]_0^{\Large\frac{\pi}{2}}-\int_0^{\Large\frac{\pi}{2}}$$\tan x/2dx+\int_0^{\Large\frac{\pi}{2}}\tan x/2dx$
Step 3:
On applying limits
$\large\frac{\pi}{2}$$\tan\Large\frac{\pi}{4}$$-0$
$\tan\large\frac{\pi}{4}$$=1$
$\Rightarrow\large\frac{\pi}{2}$
answered Nov 12, 2013 by sreemathi.v
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