Mathematics

# The value of the integral $\displaystyle \int_{0}^{\pi /2}\dfrac{\sin x+\cos x}{e^{x-\frac{\pi }{4}}+1}dx$ is

$1$

##### SOLUTION
The given integral
$I=\displaystyle \int_{0}^{\pi /2}\dfrac{\sin x+\cos x}{e^{x-\frac{\pi }{4}}+1}dx$

$\displaystyle =\int _{ 0 }^{ \pi /2 } \dfrac {\sin\left( \dfrac { \pi }{ 2 } -x \right) +\cos\left( \dfrac { \pi }{ 2 } -x \right) }{ e^{ \frac { \pi }{ 2 } -x-\frac { \pi }{ 4 } }+1 } dx$

$\displaystyle =\int _{ 0 }^{ \pi /2 } \dfrac { \cos { x }+\sin { x } }{ { e }^{ -({ x }-\frac { \pi }{ 4 } ) }+1 } { d }{ x }$

i.e., $\displaystyle I=\int_{0}^{\pi /2}\dfrac{e^{\left ( x-\frac{\pi }{4} \right )}(\sin x+\cos x)}{1+e^{x-\frac{\pi }{4}}}$

Hence $\displaystyle 2I=\int_{0}^{\pi /2}\dfrac{\sin x+\cos x}{e^{x-\frac{\pi }{4}}+1}dx+\int_{0}^{\pi /2}\dfrac{e^{x-\frac{\pi }{4}}(\sin x+\cos x)}{e^{x-\frac{\pi }{4}}+1}dx$

$\displaystyle = \int_{0}^{\pi/2}(\sin {x}+\cos {x}) dx$

$=(-\cos x+\sin x)_{0}^{{\pi }/{2}}=2$

$\Rightarrow I=1$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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