(a) Evaluate ∫_0^π / 4 log (1+ tan x ) d x.

Let $\displaystyle \mathrm{I}=\int_{0}^{\pi / 4} \log (1+\tan \mathrm{x}) \mathrm{dx}$

$$\Rightarrow \mathrm{I}=\int_{$\displaystyle 0$}^{\pi / $\displaystyle 4$} \log \left($\displaystyle 1$+\tan \left(\frac{\pi}{4}-\mathrm{x}\right)\right) \mathrm{d} \mathrm{x}=\int_{$\displaystyle 0$}^{\pi / $\displaystyle 4$} \log \left(\frac{2}{$\displaystyle 1$+\tan \mathrm{x}}\right) \mathrm{d} \mathrm{x}

$$Add (i) and (ii), $\displaystyle \left.2 \mathrm{I}=\int_{0}^{\pi / 4} \log 2 \mathrm{dx}=\log 2 \cdot \mathrm{x}\right]_{0}^{\pi / 4}=\frac{\pi}{4} \log 2$

$$\Rightarrow \mathrm{I}=\frac{\pi}{8} \log $\displaystyle 2$

$$Or (b) Let $\displaystyle \mathrm{I}=\int \frac{1}{\sqrt{\sin ^{3} \mathrm{x} \cos (\mathrm{x}-\alpha)}} \mathrm{dx}=\int \frac{\operatorname{cosec}^{2} \mathrm{x}}{\sqrt{\sin \alpha+\cos \alpha \cot \mathrm{x}}} \mathrm{dx}$

$$\begin{gathered} \text { Put } \sin \alpha+\cos \alpha \cot \mathrm{x}=t \Rightarrow \operatorname{cosec}^{$\displaystyle 2$} \mathrm{x} \mathrm{d} \mathrm{x}=-\frac{1}{\cos \alpha} \mathrm{d} t \therefore \mathrm{I}=-\int \frac{1}{\cos \alpha \sqrt{t}} \mathrm{d} t=-\frac{$\displaystyle 2$ \sqrt{t}}{\cos \alpha}+c \Rightarrow \mathrm{I}=-\frac{2}{\cos \alpha} \sqrt{\sin \alpha+\cos \alpha \cot \mathrm{x}}+c \end{gathered}