#### Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 36 textbook solution.

Answer:-  $\frac{\pi^{2}}{2 \sqrt{2}}$

Hints:-  You must know the integration rules of trigonometric functions.

Given:-  $\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x \cdot d x$

Solution : $\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x \cdot d x$                                 ....(1)

Then

$\begin{gathered} I=\int_{0}^{\pi} \frac{\pi-x}{1-\sin ^{2}(\pi-x)}+\cos ^{7}(\pi-x) \cdot d x \\ \quad I=\int_{0}^{\pi} \frac{\pi-x}{1+\sin ^{2} x}-\cos ^{7} x \cdot d x \end{gathered}$              ....(2)

\begin{aligned} &\left.2 I=\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x+\frac{\pi-x}{1+\sin ^{2} x}-\cos ^{7} x\right) \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{1+\sin ^{2} x} d x \end{aligned}

Dividing numerator and denominator by $\cos^{2} x$

\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} x}{\sec ^{2} x+\tan ^{2} x} \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} x}{1+2 \tan ^{2} x} \cdot d x \\ &2 I=\pi \int_{0}^{\pi / 2} \frac{\sec ^{2} x}{1+2 \tan ^{2} x} \cdot d x \end{aligned}                                          $\left[\because \int_{0}^{2 a} f(x) d x=\int_{0}^{2} \int_{0}^{a} f(x)\right]$

Put $\tan x = z$

Then $\sec^{2}x.dx=dz\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \text{when}x\rightarrow 0,z\rightarrow 0$

$x=\frac{\pi}{2},z\rightarrow \infty$

\begin{aligned} &2 I=2 \pi \int_{0}^{\infty} \frac{d z}{1+(\sqrt{2} z)^{2}} \\ &\left.2 I=2 \pi \times \frac{\tan ^{-1} \sqrt{2} z}{\sqrt{2}}\right]_{0}^{\infty} \end{aligned}

\begin{aligned} &I=\frac{\pi}{\sqrt{2}} \times\left(\tan ^{-1} \infty-\tan ^{-1} 0\right) \\ &I=\frac{\pi}{\sqrt{2}} \times\left(\frac{\pi}{2}-0\right) \\ &I=\frac{\pi^{2}}{2 \sqrt{2}} \end{aligned}