Question

Evaluate the following:π4∫−π4log(sinx+cosx)dx

Answer 1

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Answer 2

Answer: \begin{aligned}

\text { Let } I &=\int_{-\pi / 4}^{\pi / 4} \log (\sin x+\cos x) d x \\

&=\int_{-\pi / 4}^{\pi / 4} \log \left[\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)\right] d x \\

&=\int_{-\pi / 4}^{\pi / 4} \log \left[\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right] d x

\end{aligned}

\begin{aligned}

&\text { Put } x+\frac{\pi}{4}=t\\

&\begin{aligned}

I &=\int_{0}^{\pi / 2} \log [\sqrt{2} \sin t] d t \\

&=\int_{0}^{\pi / 2}[\log \sqrt{2}+\log \sin t] d t=\log \sqrt{2} \int_{0}^{\pi / 2} 1 d t+\int_{0}^{\pi / 2} \log \sin t d t \\

&=\frac{\pi}{2} \log \sqrt{2}-\frac{\pi}{2} \log 2=\frac{\pi}{4} \log 2-\frac{\pi}{2} \log 2=-\frac{\pi}{4} \log 2

\end{aligned}

\end{aligned}