(a)OR(b)

Find : \[\int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x \]

$$\begin{aligned} & I=\int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x \\ & =\int \frac{2+2 \sin x \cos x}{2 \cos ^{2} x} e^{x} d x \\ & =\int\left(\sec ^{2} x+\tan x\right) e^{x} d x \\ & =e^{x} \cdot \tan x+c \end{aligned} \begin{aligned} I & =\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin x+\cos x} d x \\ & =\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \dfrac{\pi}{4} \sin x+\sin \dfrac{\pi}{4} \cos x} d x \\ & =\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{4}} \frac{1}{\sin \left(x+\dfrac{\pi}{4}\right)} d x==\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{4}} \operatorname{cosec}\left(x+\frac{\pi}{4}\right) d x \\ & =\frac{1}{\sqrt{2}}\left[\log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right]_{0}^{\frac{\pi}{4}}\right. \\ & =\frac{1}{\sqrt{2}} \log (\sqrt{2}+1) \text { or }-\frac{1}{\sqrt{2}} \log (\sqrt{2}-1) \end{aligned} $$