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#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 27 maths.

Answer: $\frac{\pi}{4}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{\pi}\frac{1}{5+3\cos x}dx$

Solution: $I=\int_{0}^{\pi}\frac{1}{5+3\cos x}dx$

Put $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} d x$

$I=\int_{0}^{\pi}\frac{1}{5+3\cos x}dx$

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

\begin{aligned} &=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{8+2 \tan ^{2} \frac{x}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right] \\ &=\int_{0}^{\pi} \frac{1}{2} \frac{\sec ^{2} \frac{x}{2}}{4+\tan ^{2} \frac{x}{2}} d x \\ &=\frac{1}{2} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{2^{2}+\tan ^{2} \frac{x}{2}} d x \end{aligned}

Put $\tan \frac{x}{2}=t$

$\sec^2 \frac{x}{2}.\frac{1}{2}dx=dt$

$\sec^2 \frac{x}{2}dx=2dt$

When x=0  then t=0  and $x=\pi$  then$t=\infty$

\begin{aligned} &I=\frac{1}{2} \int_{0}^{\infty} \frac{1}{2^{2}+t^{2}} 2 d t \\ &=2 \times \frac{1}{2} \int_{0}^{\infty} \frac{1}{2^{2}+t^{2}} d t \\ &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]_{0}^{\infty} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{2}=\infty, \tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 0=0\right] \\ &=\frac{\pi}{4} \end{aligned}