INTEGRALS — Maths STD 12 Science — Question

$\Rightarrow\ \ \text{I}=\int^{\frac{\pi}{4}}\limits_{0}\log\bigg[1+\tan\bigg(\frac{\pi}{4}-\text{x}\bigg)\bigg]\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$

$\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{4}}\log\bigg[1+\frac{1-\tan\text{x}}{1+\tan\text{x}}\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{4}}\log\bigg[\frac{1+\tan\text{x}+1-\tan\text{x}}{1+\tan\text{x}}\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{4}}\log\bigg[\frac{2}{1+\tan\text{x}}\bigg]\text{dx}$

Adding eq. (i) and (ii),

$21=\int\limits_{0}^{\frac{\pi}{4}}\bigg[\log(1+\tan\text{x})+\log\bigg(\frac{2}{1+\tan\text{x}}\bigg)\bigg]\text{dx}=\int\limits_{0 }^{\frac{\pi}{4}}\bigg[\log(1+\tan\text{x)}\bigg(\frac{2}{1+\tan\text{x}}\bigg)\bigg]\text{dx}$

$\Rightarrow\ \ \ 21=\int\limits_{0}^{\frac{\pi}{4}}\big[\log2\big]\text{dx}=(\log2)\ \text{(x)}^{\frac{\pi}{4}}_{0}=\frac{\pi}{4}\log2$

$\Rightarrow\ \ \text{I}=\frac{\pi}{8}\log2$