MCQ
$\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x\,\,dx}}{{\sin x + \cos x}} = ........} $
  • A
    $\sqrt 2 \,\log \left( {\sqrt 2 - 1} \right)$
  • B
    $\sqrt 2 \,\log \left( {\sqrt 2 + 1} \right)$
  • $\frac{1}{{\sqrt 2 }}\,\log \left( {\sqrt 2 + 1} \right)$
  • D
    $\frac{1}{{\sqrt 2 }}\,\log \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}}$

Answer

Correct option: C.
$\frac{1}{{\sqrt 2 }}\,\log \left( {\sqrt 2 + 1} \right)$
$I=\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x\,\,dx}}{{\sin x + \cos x}} } ..........(1)$
$= \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}(\frac{\pi}{2}-x)\,\,dx}}{{\sin(\frac{\pi}{2}- x )+ \cos (\frac{\pi}{2}-x)}}}$
$=\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x\,\,dx}}{{\cos x + \sin x}} }\ ..........(2)$
સમી. $(1)$ અને $(2)$ નો સરવાળો કરતા,
$2I=\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x \,\, + \cos^2x}}{{\sin x + \cos x}} }dx$
$=\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{1}{2}sinx+\frac{1}{\sqrt{2}}cosx}dx$
$ =\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}} \frac{1}{\cos(x-\frac{\pi}{4})} dx$
$=\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}  \sec(x-\frac{\pi}{4}) dx $
$\frac{1}{\sqrt{2}}[\log|\sec(x-\frac{\pi}{4})+ \tan(x-\frac{\pi}{4})|]_0^{\frac{\pi}{2}}$
$ \frac{1}{\sqrt{2}}\log[\frac{\sqrt{2}+1}{\sqrt{2}-1}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}] $
$2I=\frac{2}{\sqrt{2}}\log(\sqrt{2}+1)$
$I=\frac{1}{\sqrt{2}}\log(\sqrt{2}+1)$

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