(a) Find: ∫ x √(1+2 x) dx; OR (b) Evaluate: ∫_0^(π)/4^2 (sin √ x)/(√ x) d x

$$\begin{aligned} & 1+2 \mathrm{x}=\mathrm{t}^{2} \\ & 2 d \mathrm{x}=2 \mathrm{t} d \mathrm{t} \\ & \frac{1}{2} \int\left(\mathrm{t}^{4}-\mathrm{t}^{2}\right) d \mathrm{t}=\frac{1}{2}\left[\frac{\mathrm{t}^{5}}{5}-\frac{\mathrm{t}^{3}}{3}\right]+C \\ & \qquad=\frac{(1+2 \mathrm{x})^{5}}{10}-\frac{(1+2 \mathrm{x})^{\frac{3}{2}}}{6}+C \end{aligned} $$ $\displaystyle \int_{0}^{\frac{\pi^{2}}{4}} \frac{\sin \sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}} d \mathrm{x}$ Put $\displaystyle \sqrt{\boldsymbol{\mathrm{x}}}=\boldsymbol{\mathrm{t}} \Rightarrow \boldsymbol{d} \boldsymbol{\mathrm{x}}=\mathbf{2} \boldsymbol{\mathrm{t}} \boldsymbol{d} \boldsymbol{\mathrm{t}} 2 \int_{0}^{\frac{\pi}{2}} \sin \mathrm{t} d \mathrm{t}=2[-\cos \mathrm{t}]_{0}^{\frac{\pi}{2}}$ $$\text { = } 2 $$