#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 1 Maths Textbook Solution.

Answer: $\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$

Given: $\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$

Hint: Use substitution method

Solution: $\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$

Dividing numerator and denominator by $\cos ^{2} x$

\begin{aligned} &=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \cos ^{2} x}{\cos ^{2} x}+\frac{9 \sin ^{2} x}{\cos ^{2} x}} d x \\ &=\int \frac{\sec ^{2} x d x}{4+9 \tan ^{2} x} \end{aligned}

$Now,Let$

\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}                                                                        (Differentiating w.r.t to x)

$\text { So, } \int \frac{d t}{4+9 t^{2}}$

$=\frac{1}{9} \int \frac{d t}{\frac{4}{9}+t^{2}}$

$=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1}\left(\frac{t}{\frac{2}{3}}\right)\right]+c$                                            $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$

$=\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$                                                $[\tan x=t]$