#### Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 4 maths textbook solution.

$I =\frac{1}{\sqrt{34}}\left [ \tan^{-1} \frac{5\sqrt{34}}{28}\right ]$

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

Given: $\int_{0}^{\frac{\pi}{2}} \frac{1}{5 \cos x+3 \sin x} d x$

Solution:$\int_{0}^{\frac{\pi}{2}} \frac{1}{5 \cos x+3 \sin x} d x$

Putting $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan 2 \frac{x}{2}}$

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

$=\int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{5-5 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$

$=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} \frac{x}{2}}{5+6 \tan \frac{x}{2}-5 \tan \frac{2}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right]$

$I=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+6 \tan \frac{x}{2}-5 \tan \frac{2}{2}} d x$

Let$\tan \frac{x}{2}=t$

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

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

When x=0  then $t=\tan0=0$  and when$x=\frac{\pi}{2}$  then $t=\tan \frac{\pi}{4}=1$

$I=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+6 \tan \frac{x}{2}-5 \tan \frac{2}{2}} d x$

\begin{aligned} &=\int_{0}^{1} \frac{1}{5+6 t-5 t^{2}} 2 d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{1+\frac{6}{5} t-t^{2}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left(t^{2}-\frac{6}{5} t-1\right)} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[t^{2}-2 t \frac{3}{5}+\left(\frac{3}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}-1\right]} d t \end{aligned}

\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[\left(t-\frac{3}{5}\right)^{2}-\frac{9}{25}-1\right]} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[\left(t-\frac{3}{5}\right)^{2}-\left(\frac{9+25}{25}\right)\right]} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[\left(t-\frac{3}{5}\right)^{2}-\left(\frac{34}{25}\right)\right]} d t \end{aligned}

\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(\frac{\sqrt{34}}{5}\right)^{2}+\left(t-\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(\frac{\sqrt{34}}{5}\right)^{2}+\left(t-\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5}\left[\frac{1}{\frac{\sqrt{34}}{5}} \tan ^{-1} \frac{\left(t-\frac{3}{5}\right)}{\left(\frac{\sqrt{34}}{5}\right)}\right]_{0}^{1} \\ &{\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]} \end{aligned}

\begin{aligned} &=\frac{2}{5} \frac{5}{\sqrt{34}}\left[\tan ^{-1} \frac{\left(1-\frac{3}{5}\right)}{\left(\frac{\sqrt{34}}{5}\right)}-\tan ^{-1} \frac{\left(0-\frac{3}{5}\right)}{\left(\frac{\sqrt{34}}{5}\right)}\right] \\ &=\frac{1}{\sqrt{34}}\left[\tan ^{-1} \frac{(2)}{(\sqrt{34})}-\tan ^{-1} \frac{(-3)}{(\sqrt{34})}\right] \end{aligned}

$=\frac{1}{\sqrt{34}}\left[\tan ^{-1} \frac{\left(\frac{(2)}{(\sqrt{34})}-\frac{(-3)}{(\sqrt{34})}\right)}{1+\frac{(2)}{(\sqrt{34})} \frac{(-3)}{(\sqrt{34})}}\right]$

\begin{aligned} &=\frac{1}{\sqrt{34}}\left[\tan ^{-1} \frac{\left(\frac{(2+3)}{(\sqrt{34})}\right)}{\frac{34-6}{34}}\right] \\ &=\frac{1}{\sqrt{34}} \mid \tan ^{-1} \frac{\frac{5}{(\sqrt{34})}}{\frac{28}{34}} \end{aligned}

$I =\frac{1}{\sqrt{34}}\left [ \tan^{-1} \frac{5\sqrt{34}}{28}\right ]$