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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 3 Maths Textbook Solution.

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Answer: \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x+1}{\sqrt{3}}\right)+c

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

Hint: You must know about substitution method

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

\begin{aligned} &=\int \frac{2}{2+2 \sin x \cos x} d x \\ &=\int \frac{1}{1+\sin x \cos x} d x \end{aligned}

Dividing the numerator and denominator by \cos ^{2} x

\begin{aligned} &=\int \frac{\sec ^{2} x}{\sec ^{2} x+\tan x} d x \\ &=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+\tan x} d x \end{aligned}


\begin{aligned} &\tan x=t \\ &\sec ^{2} d x=d t \end{aligned}                                                (Differentiating w.r.t x)

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

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

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


=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c                                                    \left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]


\begin{aligned} &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)+c \\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x+1}{\sqrt{3}}\right)+c \end{aligned}

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