(a)OR(b)

Evaluate : $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos 2 \mathrm{x}}{1+\cos 2 \mathrm{x}} \mathrm{~d} \mathrm{x}$

$$\begin{aligned} & \mathrm{I}=\int_{-\pi / 4}^{\pi / 4} \frac{\cos 2 \mathrm{x}}{1+\cos 2 \mathrm{x}} \mathrm{dx}=2 \int_{0}^{\pi / 4} \frac{\cos 2 \mathrm{x}}{1+\cos 2 \mathrm{x}} \mathrm{dx} \\ & =2 \int_{0}^{\pi / 4}\left(1-\frac{1}{1+\cos 2 \mathrm{x}}\right) \mathrm{d} \mathrm{x} \\ & =2 \int_{0}^{\pi / 4}\left(1-\frac{1}{2 \cos ^{2} \mathrm{x}}\right) \mathrm{d} \mathrm{x} \\ & =2 \int_{0}^{\pi / 4}\left(1-\frac{1}{2} \sec ^{2} \mathrm{x}\right) \mathrm{d} \mathrm{x} \\ & =\left.(2 \mathrm{x}-\tan \mathrm{x})\right|_{0} ^{\pi / 4} \\ & =\left(\frac{\pi}{2}-1\right) \end{aligned} Let $\displaystyle \mathrm{I}=\int \mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{x}^{5}+2 \mathrm{x}^{3}\right) \mathrm{d} \mathrm{x}$ Put $\displaystyle \mathrm{x}^{2}=\mathrm{t}$ so that $\displaystyle 2 \mathrm{xdx}=\mathrm{dt} \therefore \mathrm{I}=\frac{1}{2} \int \mathrm{e}^{\mathrm{t}}\left(\mathrm{t}^{2}+2 \mathrm{t}\right) \mathrm{d} \mathrm{t}$ \begin{aligned} & =\frac{1}{2} \mathrm{e}^{\mathrm{t}_{\mathrm{t}} 2}+\mathrm{C} \\ & =\frac{1}{2} \mathbf{e}^{\mathbf{x}^{2}}\left(\mathbf{x}^{4}\right)+\mathbf{C} \end{aligned} $$