(a) Evaluate: ∫_0^((π)/4) (sin x + cos x)/(9+16 sin 2 x) dx; OR (b) Evaluate: ∫_0^((π)/2) sin 2 x tan^-1( sin x ) d x

Let $\displaystyle \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \frac{\sin \mathrm{x}+\cos \mathrm{x}}{9+16 \sin 2 \mathrm{x}} d \mathrm{x}$ Let $\displaystyle \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \boldsymbol{\operatorname { s i n }} 2 \mathrm{x} \boldsymbol{\operatorname { t a n }}^{-1}(\boldsymbol{\operatorname { s i n }} \mathrm{x}) d \mathrm{x}$ Put $\displaystyle \sin \mathrm{x}=\mathrm{t}$ so that $\displaystyle \cos \mathrm{xdx}=\mathrm{dt}$ $\displaystyle \mathrm{I}=2 \int_{0}^{\mathbf{1}} \boldsymbol{\mathrm{t}} \boldsymbol{\operatorname { t a n }}^{-\mathbf{1}} \boldsymbol{\mathrm{t}} \boldsymbol{d} \boldsymbol{\mathrm{t}}$ $\displaystyle =2\left[\tan ^{-1} \mathrm{t}\left(\frac{\mathrm{t}^{2}}{2}\right)-\frac{1}{2} \int \frac{\mathrm{t}^{2}}{1+\mathrm{t}^{2}} d \mathrm{t}\right]_{0}^{1}$ $\displaystyle =2\left[\left(\frac{\mathrm{t}^{2}}{2}\right) \tan ^{-1} \mathrm{t}-\frac{1}{2} \mathrm{t}+\frac{1}{2} \tan ^{-1} \mathrm{t}\right]_{0}^{1}$ $\displaystyle =2\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi}{2}-1$