Question

The value of the integral ${\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\left(\frac{\sin x+\cos x}{e^{x-}+1}\right)dx$

• A. $0$
• B. $1$
• C. $2$
• D. $noneofthese$

Answer 1

The correct option is A $0$

$\begin{array}{c}Wehave\\ I={\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\left(\frac{\sin x+\cos x}{e^{x-}+1}\right)\\ {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\\ {\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\frac{\sin \left(\frac{\pi }{2}-x\right)+\cos \left(\frac{\pi }{2}-x\right)}{e^{\frac{\pi }{4}}+1}dx\\ ={\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\frac{\sin x+\cos x\left(e^{x-\frac{\pi }{4}}\right)}{e^{x-\frac{\pi }{4}}+1}=1\\ again\\ 2I={\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\left(\sin x+\cos x\right)dx\\ I=\frac{1}{2}{\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\sin xdx+\frac{1}{2}{\int }_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\cos xdx\\ =\frac{1}{2}{\left[-\cos x\right]}_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}+\frac{1}{2}{\left[\sin x\right]}_{\frac{5\pi }{4}}^{-\frac{3\pi }{4}}\\ =\frac{1}{2}\left(0+0\right)=0\\ Hence,optionAisthecorrectanswer.\end{array}$