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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 2 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 2 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c

Given:\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x

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

Solution:

\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x

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

\Rightarrow \int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \sin ^{2} x}{\cos ^{2} x}+5 \frac{\cos ^{2} x}{\cos ^{2} x}} d x

=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x

Let \begin{gathered} t=\tan x \\ d t=\sec ^{2} x d x \end{gathered}                                        (Differentiating w.r.t to x)

Now, \int \frac{1}{4 t^{2}+5} d t

=\frac{1}{4} \int \frac{1}{t^{2}+\frac{5}{4}} 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}{4} \times \frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right)+c

=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c

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