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Q. The value of the definite integral $\int\limits_{\frac{-3 \pi}{4}}^{\frac{5 \pi}{4}} \frac{\cos x+\sin x}{1+e^{x-\frac{\pi}{4}}} d x$ equals

Integrals

Solution:

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