$\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$

Question

Aicharya · Accepted Answer

Let $\begin{aligned} & I=\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}ight) d x \ &=\int_0^{\frac{\pi}{4}} \log (1+	an 	heta) \mathrm{d} 	heta \ &=\int_0^{\frac{\pi}{4}} \log \left[1+	an \left(\frac{\pi}{4}-	hetaight)ight] \mathrm{d} 	heta \ & \cdots \cdot\left[\because \int_0^a f(x) \mathrm{d} x=\int_0^a \mathrm{f}(\mathrm{a}-x) \mathrm{d} xight]\end{aligned}$ $\begin{aligned} & =\int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-	an 	heta}{1+	an 	heta}ight) d 	heta \ & =\int_0^{\frac{\pi}{4}} \log 2 d 	heta-\int_0^{\frac{\pi}{4}} \log (1+	an 	heta) d 	heta \ 	herefore \quad & 2 I=\int_0^{\frac{\pi}{4}} \log 2 d 	heta \Rightarrow I=\frac{\log 2}{2}[	heta]_0^{\pi / 4}=\frac{\pi}{8} \log 2\end{aligned}$

$\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$

Solution