Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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)

best_answer

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}

I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}-2 \cos \alpha \tan \frac{x}{2}+1} d x

Let

\begin{aligned} &\tan \frac{x}{2}=t \\ & \end{aligned}

\Rightarrow>\sec ^{2} \frac{x}{2} d x=2 d t \\

x=0 \rightarrow t=\tan 0=0 \\

x=x \rightarrow t=\tan \frac{\pi}{2}=\infty

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

I=\pi \int_{0}^{\infty} \frac{d t}{(t-\cos \alpha)^{2}+\left(1-\cos ^{2} x\right)} \\

I=\pi \int_{0}^{\infty} \frac{d t}{\sin ^{2} \alpha+(t-\cos x)^{2}}

\begin{aligned} &I=\pi-\frac{1}{\sin \alpha} \cdot \tan ^{-1}\left(\frac{t-\cos \alpha}{\sin \alpha}\right)_{0}^{\infty} \\ & \end{aligned}

I=\frac{\pi}{\sin \alpha}\left(\tan ^{-1} \infty-\tan ^{-1}\left(\frac{-\cos \alpha}{\sin \alpha}\right)\right) \\

I=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}-\tan ^{-1}(-\cot \alpha)\right)

\begin{aligned} &=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\tan ^{-1}(\cot \alpha)\right) \\ & \end{aligned}

=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\tan ^{-1} \tan \left(\frac{\pi}{2}+\alpha\right)\right) \\

=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\left(\frac{\pi}{2}+\alpha\right)\right)

= \frac{\pi}{\sin \alpha}[\pi+\alpha]

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Latest Question