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

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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}


\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]


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