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#### Need Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 15 Maths Textbook Solution.

Answer : $\frac{1}{5} \log \left|\frac{\tan \frac{x}{2}+2}{\tan \frac{x}{2}-3}\right|+C$

Hint: To solve this question we have to use partial equation

Given:  $\int \frac{1}{5+7 \cos x+\sin x} d x$

Solution :

$I=\int \frac{1}{5+7\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$                                                 $\sin 2 \theta=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \cos 2 \theta=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

\begin{aligned} &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+7\left(1-\tan ^{2} \frac{x}{2}\right)+2 \tan \frac{x}{2}} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+7-7 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}} d x \end{aligned}

\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{12-2 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}} d x\\ &=\frac{1}{2} \int \frac{\sec ^{2} \frac{x}{2}}{6-\tan ^{2} \frac{x}{2}+\tan \frac{x}{2}} d x \end{aligned}

\begin{aligned} &\text { Put } \tan \frac{x}{2}=t \\ &\ \sec ^{2} \frac{x}{2}\cdot\frac{1}{2} d x=d t \\ &I=\int \frac{d t}{6-t^{2}+t} \\ &=\int \frac{-d t}{t^{2}-t-6} \\ &=\int \frac{-d t}{t^{2}-3 t+2 t-6} \end{aligned}

\begin{aligned} &=\int \frac{-d t}{t(t-3)+2(t-3)} \\ &=-\frac{1}{5} \int \frac{(3+2) d t}{(t-3)(t+2)} \\ &=-\frac{1}{5} \int \frac{[(t+2)-(t-3)] d t}{(t-3)(t+2)} \\ &=-\frac{1}{5} \int \frac{d t}{(t-3)}+\frac{1}{5} \int \frac{d t}{(t+2)} \end{aligned}

\begin{aligned} &=\frac{1}{5}[\log |t+2|-\log |t-3|]+C \\ &=\frac{1}{5} \log \left|\frac{t+2}{t-3}\right|+C \\ &=\frac{1}{5} \log \left|\frac{\tan \frac{x}{2}+2}{\tan \frac{x}{2}-3}\right|+C \end{aligned}