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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 10 Maths Textbook Solution.

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Answer: \frac{1}{2} \log \left|\frac{\tan x}{\tan x+2}\right|+c

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

Hint: Use the formula of sin 2x and then apply substitution method

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

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

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

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


t=\tan x

d t=\sec ^{2} x d x                                          (Differentiate w.r.t x)

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

\begin{aligned} &=\int \frac{1}{t^{2}+2 t+(1)^{2}-(1)^{2}} d t \\ &=\int \frac{1}{(t+1)^{2}-(1)^{2}} d t \end{aligned}\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right]

=\frac{1}{2} \log \left|\frac{t+1-1}{t+1+1}\right|+c

\begin{aligned} &=\frac{1}{2} \log \left|\frac{t}{t+2}\right|+c \\ &=\frac{1}{2} \log \left|\frac{\tan x}{\tan x+2}\right|+c \end{aligned}

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