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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 5 Maths Textbook Solution.

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

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

Hint: Use the formula of  \sec ^{2} x  and then apply substitution method

Solution: \int \frac{1}{1+3 \sin ^{2} x} d x

Divide numerator and denominator by \cos ^{2} x

=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\cos ^{2} x}+\frac{3 \sin ^{2} x}{\cos ^{2} x}} d x

=\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x} d x

=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+3 \tan ^{2} x} d x

\left(\sec ^{2} x=1+\tan ^{2} x\right)

=\int \frac{\sec ^{2} x}{1+4 \tan ^{2} x} d x


\mathrm{t}=\tan x                                                                    (Differentiating w.r.t x)

d t=\sec ^{2} x d x

  =\int \frac{1}{1+4 t^{2}} d t

 =\frac{1}{4} \int \frac{1}{\left(\frac{1}{2}\right)^{2}+t^{2}} d t

=\frac{1}{4} \times \frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{t}{\frac{1}{2}}\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}{2} \tan ^{-1}(2 \tan x)+c


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