#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (iii)

Answer: $\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+\mathrm{c}$

Hint: Substitute   $\tan \frac{x}{2}=\mathrm{t}$

Given: $\int \frac{3+2 \cos x+4 \sin x}{2 \sin x+\cos x+3} d x$

Explanation:

$\text { Let, } I=\int \frac{3+4 \sin x+2 \cos x}{3+2 \sin x+\cos x} d x$

$=\int \frac{6+4 \sin x+2 \cos x-3}{3+2 \sin x+\cos x} \mathrm{~d} \mathrm{x}$

$=\int \frac{6+4 \sin x+2 \cos x}{3+2 \sin x+\cos x} \mathrm{dx}-\int \frac{-3}{3+2 \sin x+\cos x} \mathrm{dx}$

$=\int \frac{2(3+2 \sin x+\cos x)}{(3+2 \sin x+\cos x)} d x-\int \frac{-3}{3+2 \sin x+\cos x} d x$

$=\int 2 d x-3 \int \frac{1}{3+2 \sin x+\cos x} d x$

substituting , $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \text { and } \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$=2 x-3 \int \frac{1}{3+2 \times\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$

$=2 x-3 \int \frac{1+\tan ^{2} \frac{x}{2}}{3\left(1+\tan ^{2} \frac{x}{2}\right)+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x$

$=2 x-3 \int \frac{\sec ^{2} \frac{x}{2}}{2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+4} d x$

Substitute, $\tan \frac{x}{2}=\mathrm{t}$

$\Rightarrow \operatorname{Sec}^{2} \frac{x}{2}\cdot \frac{1}{2} \mathrm{dx}=\mathrm{dt}$

\begin{aligned} &I=2 x-3 \int \frac{2}{2 t^{2}+4 t+4} d t \\ &I=2 x-3 \int \frac{2}{2\left(t^{2}+2 t+2\right)} d t \\ &I=2 x-3 \int \frac{1}{(t+1)^{2}+1} d t \end{aligned}

$\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}(\mathrm{t}+1)+\mathrm{c} \quad\left[\because \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c\right]$

$\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+c$