Mathematics

# $\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { \sin { ^{ 2 }x } }{ \sin { x } +\cos { x } } dx } =-\dfrac { 1 }{ 2\sqrt { 2 } } \log { \left(-2 \sqrt { 2 } +3\right) }$

##### SOLUTION
$\int_{\pi/2}^{0} \dfrac{\sin^{2} x}{\sin x + \cos x} dx$
$\int_{\pi/2}^{0} \dfrac{ (1- \cos 2x)/2}{\sin x+ \cos x} dx$
$= \underset { I_1 }{ \dfrac{1}{2} \int \dfrac{1}{\sin x + \sin x }dx } - \underset { I_2 }{ \int \dfrac{\cos 2x}{\sin x+ \cos x} dx }$
$I_{1} \Rightarrow \int \dfrac{1}{\sin x + \cos x} dx$ is $\tan \dfrac{x}{2}$
$\int \dfrac{2}{1+2\mu -\mu^{2}} d\mu$
$= 2 \int \dfrac{1}{- (\mu-1)^{2}+2}$
$=\dfrac{2}{\sqrt{2} } \left( \dfrac{In | (\mu-1)/\sqrt{2}-1 |}{2} - \dfrac{In | (\mu-1)/\sqrt{2} | -1}{2} \right)$
$=\sqrt{2} \left[ \dfrac{1}{2 In \left| \left( \dfrac{\tan x/2-1}{\sqrt{2}}+1 \right) \right|}- \dfrac{1}{2} \ In \left| \tan \dfrac{x-1}{\sqrt{2}} \right| \right]$
$I_{2} \Rightarrow \int \dfrac{\cos 2x}{\sin x + \cos x} dx$
$= \int \dfrac{\cos^{2} x- \sin^{2} x}{\sin x+ \cos x} dx = \int (\cos x- \sin x) dx$
$= \sin x + \cos x$
$\Rightarrow \dfrac{\sqrt{2}}{2}\ In \left| \dfrac{\tan x+1}{\sqrt{2}} \right|- \ In \left| \dfrac{\tan x/2 -1}{\sqrt{2}} \right|- \sin x +\cos$
$I \Rightarrow \dfrac{-1}{2} - \dfrac{1}{2} \left( \dfrac{In \left( -\dfrac{1}{\sqrt{2}+1} \right)- \ In \left( \dfrac{1}{\sqrt{2}+1} \right)}{\sqrt{2}} \right)-1$
$I= -\dfrac{In (3-2 \sqrt{2})}{2 \sqrt{2}}$
$=- \dfrac{1}{2 \sqrt{2}} In (3-2 \sqrt{2})$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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