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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 6 Maths Textbook Solution.

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Answer: \frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c

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

Hint: Divide Numerator and Denominator by \cos ^{2} x and then apply substitution method.

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

On dividing Numerator and Denominator by \cos ^{2} x

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

=\int \frac{\sec ^{2} x}{3\left(1+\tan ^{2} x\right)+2} d x

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

=\int \frac{\sec ^{2} x}{5+3 \tan ^{2} x} d x


t=\tan \: \: x                                            (Differentiating w.r.t x)

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

Now,\int \frac{1}{5+3 t^{2}} d t

=\frac{1}{3} \int \frac{1}{\left(\left(\sqrt{\frac{5}{3}}\right)^{2}+t^{2}\right)} d t

=\frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{3}}}\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}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c


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