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Name: Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 46
Uploaded: Tue, 2021-08-24 17:09
Description: Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 46

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 46

Answers (1)

Answer: $\frac{\pi}{\sin \alpha}[\pi+\alpha]$

Hint: To solve this equation we convert cos and sin into tan

Given: $\frac{\pi}{\sin \alpha}[\pi+\alpha]$

Solution: $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

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

$I =\int_{0}^{\pi} \frac{\pi-x}{1+\cos \alpha \sin x} d x \\$

$I =\int_{0}^{\pi} \frac{\pi}{1+\cos \alpha \sin x} d x-\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x$

Let

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

$2 I =\int_{0}^{\pi} \frac{\pi}{1+\cos x \cdot \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$

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