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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 37 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 37 textbook solution.

Answers (1)

Answer:-  \frac{\pi(\pi / 2-\alpha)}{\cos \alpha}

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

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

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

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

              \begin{aligned} &2 I=\int_{0}^{\pi} \frac{\pi}{1+\sin \alpha \sin x} \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{\sin \alpha \sin x+1} d x \end{aligned}

Substituting \sin x=\frac{2 \tan x / 2}{1+\tan ^{2} x / 2}

               \begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{1+\tan ^{2} x / 2}{1+\tan ^{2} x / 2+\sin \alpha+2 \tan ^{2} x / 2} d x \\ &I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} x / 2}{1+\tan ^{2} x / 2+\sin \alpha+2 \tan x / 2} d x \end{aligned}

Let  \tan \frac{x}{2}=t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { Also when } x \rightarrow 0, t \rightarrow 0

       \sec ^{2} \frac{x}{2}=2 d t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad x \rightarrow \pi, t \rightarrow \infty

            \begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{2 d t}{t^{2}+2 t \sin \alpha+1} \\ &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{1}{(t+\sin \alpha)^{2}+\cos ^{2} \alpha} d t \\ &I=\frac{\pi}{\cos \alpha}\left[\tan ^{-1}\left(\frac{1+\sin \alpha}{\cos \alpha}\right)\right]_{0}^{\infty} \end{aligned}

           \begin{aligned} &I=\frac{\pi}{\cos \alpha}\left[\tan ^{-1} \infty-\tan ^{-1}(\tan \alpha)\right] \\ &I=\frac{\pi}{\cos \alpha}\left[\frac{\pi}{2}-\alpha\right] \end{aligned}

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