Question

The value of the definite integral $\underset{\frac{-3\pi }{4}}{\overset{\frac{5\pi }{4}}{\int }}\frac{\cos x+\sin x}{1+e^{x-\frac{\pi }{4}}}dx$ equals

• A. 0
• B. 1
• C. 3
• D. 4

Answer 1

Answer: (A)

Using King
$I=\underset{\frac{-3\pi }{4}}{\overset{1+e^{\frac{5\pi }{4}}}{\int }}\frac{\sin x+\cos x}{-\left(x-\frac{\pi }{4}\right)}dx;I=\underset{\frac{-3\pi }{4}}{\overset{\frac{5\pi }{4}}{\int }}\frac{(\cos x+\sin x)}{1+e^{\left(x-\frac{\pi }{4}\right)}}{e}^{\left(x-\frac{\pi }{4}\right)}dx$
${\therefore 2I=\underset{\frac{-3\pi }{4}}{\overset{\frac{5\pi }{4}}{\int }}(\cos x+\sin x)dx=(\sin x-\cos x)]}_{\frac{-3\pi }{4}}^{\frac{5\pi }{4}}=\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$
$2I=0\Rightarrow I=0$