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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 64 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 64 Maths Textbook Solution.

Answers (1)

Answer:

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

Hint

We will write \cos x \operatorname{as}\left(\frac{1-\tan ^{2} \frac{x \lambda}{2}}{1+\tan ^{2} \frac{x}{2}}\right)

Given:

\int \frac{1}{1+2 \cos x} d x

Solution:

I=\int \frac{1}{1+2 \cos x} d x

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

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

I=\int \frac{\sec ^{2} \frac{x}{2}}{3-\tan ^{2} \frac{x}{2}} d x

I=\int \frac{2 d t}{3-t^{2}}                                                \left[\because \tan \frac{\mathrm{x}}{2}=t, d t=\sec ^{2}\left(\frac{x}{2}\right)\left(\frac{1}{2}\right) d x\right]

I=2 \int \frac{d t}{(\sqrt{3})^{2}-t^{2}}                                \left[\because \sec ^{2}\left(\frac{x}{2}\right) d x=2 d t\right]

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

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

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

 

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