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Question Number 79186 by john santu last updated on 23/Jan/20

$$\underset{-1}{\stackrel{1}{\int }}\frac{\cos \left(x\right)}{1+e^{\frac{1}{x}}}dx?$$

Answered by mind is power last updated on 23/Jan/20

$$={\int }_{-1}^0\frac{cos\left(x\right)}{1+e^{\frac{1}{x}}}dx+{\int }_0^1\frac{cos\left(x\right)}{1+e^{\frac{1}{x}}}dx$$$${\int }_{-1}^0\frac{cos\left(x\right)}{1+e^{\frac{1}{x}}}dx,x=-u\Rightarrow ={\int }_0^1\frac{cos\left(u\right)}{1+e^{-\frac{1}{u}}}du$$$$\Rightarrow {\int }_{-1}^1\frac{cos\left(x\right)}{1+e^{\frac{1}{x}}}dx={\int }_0^1\frac{cos\left(x\right)}{1+e^{-\frac{1}{x}}}dx+{\int }_0^1\frac{cos\left(x\right)}{1+e^{\frac{1}{x}}}dx..E$$$$\frac{cos\left(x\right)}{1+e^{-\frac{1}{x}}}=\frac{cos\left(x\right)e^{\frac{1}{x}}}{e^{\frac{1}{x}}+1}$$$$E\iff {\int }_0^1(\frac{cos\left(x\right)e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}+\frac{cos\left(x\right)}{1+e^{\frac{1}{x}}})dx={\int }_0^{1}cos\left(x\right)dx=sin\left(1\right)$$

Commented by john santu last updated on 23/Jan/20

$$thankyousir$$

Commented by mind is power last updated on 23/Jan/20

$$y^{\prime }rewelcom$$

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