Mathematics

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


ANSWER

$$1$$


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
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