MCQ
$\int_{\,0}^{\,\pi } {\sqrt {\frac{{1 + \cos 2x}}{2}} \,dx} =$
- A$0$
- ✓$2$
- C$1$
- D$ - 1$
$I = \int_{\,0}^{\,\pi /2} {\cos x\,dx} - \int_{\,\pi /2}^{\,\pi } {\cos x\,dx} $
$= [\sin x]_0^{\pi /2} - [\sin x]_{\pi /2}^\pi $
$I = \left[ {\sin \frac{\pi }{2} - \sin 0} \right] - \left[ {\sin \pi - \sin \frac{\pi }{2}} \right] $
$=1+ 1 = 2.$
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વિધાન-$2$ : ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right]$ = ${\tan ^{ - 1}}\,1 + \,{\tan ^{ - 1}}\left( {\log {x^2}} \right)$
($1$) The value of $\frac{625}{4} p _1$ is
($2$) The value of $\frac{125}{4} p _2$ is
Give the answer or queution ($1$) and ($2$)