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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.

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

Answers (1)

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

Let

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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