#### Provide Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 6 Maths Textbook Solution.

Answer : $\tan ^{-1}\left[1+\tan \frac{x}{2}\right]+c$

Hint: To solve this statement we have to convert sinx and cosx in terms of tanx.

Given:  $\int \frac{1}{3+2 \sin x+\cos x} d x$

Solution : $I=\int \frac{1}{3+2 \sin x+\cos x} d x$

We use the formula : $\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]$

\begin{aligned} &I=\int \frac{1}{3+2 \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{3+3 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \end{aligned}

\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+4} d x \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{2\left(\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+2\right)} d x \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} t=\tan \frac{x}{2} \\ \frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \end{array}\right]} \\ &I=\int \frac{d t}{t^{2}+2 t+2} \\ &I=\int \frac{d t}{\left(t^{2}+2 t+1\right)+1} \end{aligned}

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