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  • Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 36

Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 36

Answers (1)

Answer:

\frac{\pi }{4}

Hint:

To solve this equation we convert cos in term of tan.

Given:

\int_{0}^{\pi} \frac{1}{5+3 \cos x} d x

Solution:

Let

\begin{aligned} &I=\int_{0}^{\pi} \frac{1}{5+3 \cos x} d x \\\\ &\therefore \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \end{aligned}

\cos \theta=\frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}

I=\int_{0}^{\pi} \frac{1}{5+3\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x

\begin{aligned} &I=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{8+2 \tan ^{2} \frac{x}{2}} d x \\\\ &\text { Let } \tan \frac{x}{2}=t, \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}

\begin{aligned} &=\frac{2}{2} \int_{0}^{\infty} \frac{d t}{4+t^{2}} \\\\ &=\int_{0}^{\infty} \frac{1}{(2)^{2}+t^{2}} d t \end{aligned}

\begin{aligned} &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]_{0}^{\infty} \\\\ &=\frac{1}{2}\left(\frac{\pi}{2}\right)=\frac{\pi}{4} \end{aligned}

 

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