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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 63 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 63 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{\sqrt{3}} \ln \left|\frac{\sin x-\left(\frac{\sqrt{3}}{2}\right)}{\sin x+\left(\frac{\sqrt{3}}{2}\right)}\right|+c

Hint:

To solve this given statement we will write cos²x in terms of sin²x

Given:

\int \frac{\cos x}{\frac{1}{4}-\cos ^{2} x} d x

Solution:

\int \frac{\cos x}{\frac{1}{4}-\cos ^{2} x} d x

  I=\int \frac{\cos x}{\frac{1}{4}-\left(1-\sin ^{2} x\right)} d x

I=\int \frac{\cos x}{-\frac{3}{4}+\sin ^{2} x} d x

 

\sin x=t \quad d t=\cos x d x

\therefore I=\int \frac{d t}{t^{2}-\frac{3}{4}}

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

I=\frac{1}{2\left(\frac{\sqrt{3}}{2}\right)} \ln \left|\frac{t-\left(\frac{\sqrt{3}}{2}\right)}{t+\left(\frac{\sqrt{3}}{2}\right)}\right|+c

I=\frac{1}{\sqrt{3}} \ln \left|\frac{\sin x-\left(\frac{\sqrt{8}}{2}\right)}{\sin x+\left(\frac{\sqrt{8}}{2}\right)}\right|+c

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