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Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 65  Maths Textbook Solution.

Answers (1)

Answer:

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

Hint:

To solve the above equation we will write sin x as \frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}

Given:

\int \frac{1}{1-2 \sin x} d x

Solution: 

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

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

I=\int \frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x                                        \left[\because \sin x=\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\right.

I=\int \frac{\sec ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x

I=\int \frac{2}{2} \frac{\sec ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x

I=2 \int \frac{d t}{1+t^{2}-4 t}                                \left[\because t=\tan \left(\frac{x}{2}\right), \frac{d t}{d x}=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right), d t=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x\right]

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

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

I=2 \int \frac{d u}{u^{2}-(\sqrt{3})^{2}}                            [\because t-2=u, d t=d u]

\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right.

I=2 \times \frac{1}{2 \sqrt{3}} \log \left|\frac{u-\sqrt{3}}{u+\sqrt{3}}\right|+c

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

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

 

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