(a)OR(b)

Evaluate : $\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{x \mathrm{d} x}{1+\cos 2 x+\sin 2 x}$

$$\begin{align*} & I=\int_{0}^{\frac{\pi}{4}} \frac{x}{1+\cos 2 x+\sin 2 x} \mathrm{d} x \ldots(1) \tag{1}\\ & \text { On applying } \int_{0}^{a} f(x) \mathrm{d} x=\int_{0}^{a} f(a-x) \mathrm{d} x \\ & \text { we get } I=\int_{0}^{\frac{\pi}{4}} \frac{\dfrac{\pi}{4}-x}{1+\cos 2 x+\sin 2 x} \mathrm{d} x \ldots(2) \tag{2} \end{align*} On adding Eq. ($\displaystyle 1$) and ($\displaystyle 2$), we get $\displaystyle 2 I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{4}} \frac{1}{1+\cos 2 x+\sin 2 x} \mathrm{d} x$ I=\frac{\pi}{16} \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos ^{2} x+\sin x \cos x} \mathrm{d} x=\frac{\pi}{16} \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{2} x \mathrm{d} x}{1+\tan x} $\displaystyle I=\frac{\pi}{16}(\log |1+\tan x|)_{0}^{\frac{\pi}{4}}$ $\displaystyle I=\frac{\pi}{16} \log 2$ \begin{aligned} & I=\int \mathrm{e}^{x}\left(\frac{x}{\sqrt{1+x^{2}}}+\frac{1}{\left(1+x^{2}\right)^{\frac{3}{2}}}\right) \mathrm{d} x \\ & \operatorname{Let} f(x)=\frac{x}{\sqrt{1+x^{2}}}, f^{\prime}(x)=\frac{\sqrt{1+x^{2}}-x \dfrac{x}{\sqrt{1+x^{2}}}}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{\frac{3}{2}}} \\ & \text { On applying } \int \mathrm{e}^{x}\left(f(x)+f^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^{x} f(x)+c \\ & I=\mathrm{e}^{x} \frac{x}{\sqrt{1+x^{2}}}+c \end{aligned} $$