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Question Number 27294 by iy last updated on 04/Jan/18

$$\underset{1/e}{\stackrel{\tan x}{\int }}\frac{t}{1+t^2}dt+\underset{1/e}{\stackrel{\cot x}{\int }}\frac{1}{t(1+t^2)}dt=$$

Commented by prakash jain last updated on 05/Jan/18

$$\left(A\right)$$$${\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt=\frac{1}{2}[\ln (1+{\tan }^{2}x)-\ln (1+\frac{1}{e^2})]$$$$=\frac{1}{2}[\ln \left({\sec }^{2}x\right)-\ln (1+\frac{1}{e^2})]$$$$=\ln \left(\sec x\right)-\frac{1}{2}\ln (1+\frac{1}{e^2})$$$${\int }_{1/e}^{\cot x}\frac{1}{t(1+t^2)}dt$$$${\int }_{1/e}^{\cot x}(\frac{t}{t^2}-\frac{t}{1+t^2})dt$$$$={[\ln t-\frac{1}{2}\ln (1+t^2)]}_{1/e}^{\cot x}$$$$\left(B\right)$$$$=\ln \cot x-\frac{1}{2}\ln (1+{\cot }^{2}x)-\ln \frac{1}{e}+\frac{1}{2}\ln (1+\frac{1}{e^2})$$$$=\ln \cot x-\ln \mathrm{cosec}x-\ln \frac{1}{e}+\frac{1}{2}\ln (1+\frac{1}{e^2})$$$$A+B$$$$\ln \sec x+\ln \cot x-\ln \mathrm{cosec}x+1$$$$=\ln (\sec x\cdot \cot x)-\ln \mathrm{cosec}x+1$$$$=\ln \mathrm{cosec}x-\ln \mathrm{cosec}x+1$$$$=1$$

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