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Question Number 201292 by sonukgindia last updated on 03/Dec/23

Answered by aleks041103 last updated on 03/Dec/23

$$I=\underset{-a}{\stackrel{a}{\int }}\frac{cos\left(x\right)dx}{1+e^{\pi /x}}=\underset{a}{\stackrel{-a}{\int }}\frac{cos(-x)d(-x)}{1+e^{\pi /(-x)}}=$$$$={\int }_{-a}^a\frac{cos\left(x\right)dx}{1+e^{-\pi /x}}={\int }_{-a}^a\frac{e^{\pi /x}}{1+e^{\pi /x}}cos\left(x\right)dx$$$$\Rightarrow I+I=2I=\underset{-a}{\stackrel{a}{\int }}\frac{1+e^{\pi /x}}{1+e^{\pi /x}}cos\left(x\right)dx$$$$\Rightarrow 2I=\underset{-a}{\stackrel{a}{\int }}cos\left(x\right)dx=2sin\left(a\right)$$$$\Rightarrow I=sin\left(a\right)$$$$inthiscasea=\phi$$$$\Rightarrow I=\underset{-\phi }{\stackrel{\phi }{\int }}\frac{cos\left(x\right)dx}{1+e^{\pi /x}}=sin\left(\phi \right)$$

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