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$\Large \int\frac{x+\sin x}{1+\cos x}\normalsize dx$ is equal to\begin{array}{1 1}(A)\;\log\mid1+\cos x\mid+C & (B)\;\log\mid x+\sin x\mid+C\\(C)\;x-\tan\frac{x}{2}+C & (D)\;x\;\tan\frac{x}{2}+C \end{array}

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  • $1+\cos x=2\cos^2\large\frac{x}{2}$
  • $\sin x=2\sin\large\frac{x}{2}\normalsize \cos\large\frac{x}{2}$
  • $\int udv=uv-\int v du$
  • $\sec^2xdx=\tan x+c.$
Step 1:
Let $I=\Large \int\frac{x+\sin x}{1+\cos x}\normalsize dx$
$\sin x=2\sin \large\frac{x}{2}\normalsize \cos\large \frac{x}{2}$ and $1+\cos x=2\cos^2x$
$I=\Large \int\frac{x+2\sin \Large\frac{x}{2}\cos\Large\frac{x}{2}}{ 2\cos^2\Large\frac{x}{2}}\normalsize dx$
This can be splitted as
$\int\Large\frac{x}{2\cos^2\Large\frac{x}{2}}\normalsize dx+\int\Large\ \frac{2\sin\frac{x}{2}\cos\Large\frac{x}{2}}{\Large 2\cos^2\Large\frac{x}{2}}\normalsize dx$
But $\large\frac{1}{\cos^2\Large\frac{x}{2}}$$=\sec^2\Large\frac{x}{2}$
$I=\large\frac{1}{2}$$\int x\sec^2\large\frac{x}{2}$$dx+\int \tan\large\frac{x}{2}$$dx$------(1)
Consider $\int x\sec^2\large\frac{x}{2}$$dx$
Clearly this is of the form $u dv=uv-\int vdu.$
Step 2:
Let $u=x,$on differentiating with respect to $x$
$du=dx$ and let $dv=\sec^2xdx$
On integrating we get $v=2\tan\large\frac{x}{2}$
Now substituting for $u,v,du$ and $dv$ we get,
$\int x\sec^2\large\frac{x}{2}$$dx=x(2\tan\large\frac{x}{2})$$-2\int \tan\frac{x}{2}$$dx$
$\quad\qquad\qquad=2x\tan\frac{x}{2}-2\int \tan\large\frac{x}{2}$
Now substituting in equ (1)
$\large\frac{1}{2}$$[2x\tan\large\frac{x}{2}-$$2\int \tan\large\frac{x}{2}]$$+\int \tan\large\frac{x}{2}$$dx$
Hence the correct option is D.
answered Apr 24, 2013 by sreemathi.v

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