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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 41 Maths Textbook Solution.

Explain Solution R.D Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 41 Maths Textbook Solution.

Answers (1)

Answer:

\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|

Given:

\int \frac{1}{\sin (x-a) \sin (x-b)} d x

Hint:

Using  \sin (x-y)=\sin x \cos y-\cos x \sin y \text { and } \int \frac{1}{x} d x

Explanation:

Let I=\int \frac{1}{\sin (x-a) \sin (x-b)} d x

         =\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)} d x \ldots \ldots \text { multiplying by } \sin (b-a) \text { in num.\& den. }

         =\frac{1}{\sin (b-a)} \int \frac{\sin ((x-a)-(x-b))}{\sin (x-a) \sin (x-b)} d x                                                [\sin (x-y)=\sin x \cos y-\cos x \sin y]

         \begin{aligned} &=\frac{1}{\sin (b-a)} \int \frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)} d x\\ &=\frac{1}{\sin (b-a)}\left[\int \frac{\cos (x-b)}{\sin (x-b)} d x-\int \frac{\cos (x-a)}{\sin (x-a)}\right] d x\\ &=\frac{1}{\sin (b-a)}[\log |\sin (x-b)|-\log |\sin (x-a)|]+C \end{aligned}

         =\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|                                                            \left[\because \log a-\log b=\log \frac{a}{b}\right]

 

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