$\int_0^{\pi / 4} \log (1+ an x) d x=$

Question

$\int_0^{\pi / 4} \log (1+	an x) d x=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } I=\int_0^{\pi / 4} \log (1+	an x) d x \ & =\int_0^{\pi / 4} \log \left[1+	an \left(\frac{\pi}{4}-xight)ight] d x \ & =\int_0^{\pi / 4} \log \left[1+\left(\frac{1-	an x}{1+	an x}ight)ight] d x=\int_0^{\pi / 4} \log \left(\frac{2}{1+	an x}ight) d x \ & =\int_0^{\pi / 4}(\log 2) d x-\int_0^{\pi / 4} \log (1+	an x) d x=\int_0^{\pi / 4}(\log 2) d x-I \ & 	herefore 2 I=(\log 2)[x]_0^{\pi / 4}=(\log 2)\left(\frac{\pi}{4}ight) \Rightarrow I=\left(\frac{\pi}{8}ight) \log 2\end{aligned}$

$\int_0^{\pi / 4} \log (1+\tan x) d x=$

Solution