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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 11 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 11 maths Textbook Solution.

Answers (1)

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

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

Hint: Divide numerator and denominator by \cos ^{2} x and then use substitution method

Solution:

\begin{aligned} &\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x \\ &\left(\cos 2 x=\cos ^{2} x-\sin ^{2} x\right) \\ &=\int \frac{1}{\cos ^{2} x-\sin ^{2} x+3 \sin ^{2} x} d x \\ &=\int \frac{1}{\cos ^{2} x+2 \sin ^{2} x} d x \end{aligned}

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

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

Let

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

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

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

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

=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+c

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