#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 11 maths Textbook Solution.

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

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

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

Solution:

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

On dividing numerator and denominator by $\cos ^{2} x$, we get

$=\int \frac{\sec ^{2} x}{1+2 \tan ^{2} x} d x$

$Let$

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

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

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

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

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

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