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  • Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 27 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 27 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x+1}{\sqrt{2} \cos x-1}\right|+c

Given:

\int \frac{\sin x}{\cos 2 x} d x

Hint:

To solve the equation we will use substitution method.

Solution: 

I=\int \frac{\sin x}{\cos 2 x} d x

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

I=-\int \frac{d t}{2 t^{2}-1} \ldots[\mathrm{put} \cos \mathrm{x}=\mathrm{t},-\sin x \mathrm{dx}=\mathrm{d} \mathrm{t}]

I=\frac{1}{2} \int \frac{d t}{\frac{1}{2}-t^{2}}

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

I=-\frac{1}{2} \frac{\sqrt{2}}{2} \log \left|\frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}}\right|+c

I=\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x+1}{\sqrt{2} \cos x-1}\right|+\mathrm{C}

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