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#### Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 11 textbook solution.

Answer:-  $\frac{\pi^{2}}{4}$

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

Given:-  $\int_{0}^{\pi} \frac{x \tan x}{\sec x \cos e c x} d x$

Solution : $I=\int_{0}^{\pi} \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x} \times \frac{1}{\sin x}} d x$

$I=\int_{0}^{\pi} x \sin ^{2} x \cdot d x$                                ......(1)

\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \sin ^{2}(\pi-x) \cdot d x\\ &I=\int_{0}^{\pi}(\pi-x) \sin ^{2} x d x \end{aligned}         .......(2)

Adding (1) and (2)

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

\begin{aligned} &2 I=\frac{\pi}{2}\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi} \\ &2 I=\frac{\pi}{2}\left[\left(\pi-\frac{\sin 2 \pi}{2}\right)-\left(0-\frac{\sin 0}{2}\right)\right] \end{aligned}

\begin{aligned} &2 I=\frac{\pi}{2}[\pi] \\ &2 I=\frac{\pi^{2}}{2} \\ &I=\frac{\pi^{2}}{4} \end{aligned}

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