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  • Provide solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 15 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 15 textbook solution.

Answers (1)

Answer:- \frac{(\pi-2) \pi}{2}

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

Given:-   \int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x

Solution : \int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x

I=\int_{0}^{\pi} \frac{(\pi-x) x \sin (\pi-x)}{1+\sin (\pi-x)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]

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

\begin{aligned} &\frac{2 I}{\pi}=\int_{0}^{\pi} \frac{\sin x-\sin ^{2} x}{\cos ^{2} x} d x \\ &\frac{2 I}{\pi}=\int_{0}^{\pi} \frac{\sin x}{\cos ^{2} x} d x-\int_{0}^{\pi} \frac{\sin ^{2} x}{\cos ^{2} x} d x \\ &\frac{2 I}{\pi}=\int_{0}^{\pi} \sec x \cdot \tan x \cdot d x-\int_{0}^{\pi} \tan ^{2} x \cdot d x \end{aligned}

\begin{aligned} &\frac{2 I}{\pi}=[\sec \pi-\sec 0]-\int_{0}^{\pi} \sec ^{2} x \cdot d x+\int_{0}^{\pi} 1 \cdot d x \\ &\frac{2 I}{\pi}=[-1-1]-[\tan ]_{0}^{\pi}-[x]_{0}^{\pi} \end{aligned}

\begin{aligned} &\frac{2 I}{\pi}=[-2]-[\tan \pi-\tan 0]+\pi \\ &\frac{2 I}{\pi}=[-2]-0+\pi \\ &\therefore I=\frac{(\pi-2) \pi}{2} \end{aligned}

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