#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 2 Maths Textbook Solution.

Answer: $\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$

Given:$\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$

Hint: Divide the numerator and denominator by $\cos ^{2} x$ and then use substitution method

Solution:

$\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$

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

$\Rightarrow \int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \sin ^{2} x}{\cos ^{2} x}+5 \frac{\cos ^{2} x}{\cos ^{2} x}} d x$

$=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x$

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

Now, $\int \frac{1}{4 t^{2}+5} d t$

$=\frac{1}{4} \int \frac{1}{t^{2}+\frac{5}{4}} d t$                                         $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$

$=\frac{1}{4} \times \frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right)+c$

$=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$