Mathematics

The value of $$\int {\dfrac{{dx}}{{\sin x.\sin \left( {x + \alpha } \right)}}} $$ is equal to


SOLUTION
$$\displaystyle I=\int \dfrac{dx}{ \sin x . \sin (x+\alpha)}$$

$$\displaystyle =\dfrac{1}{\sin a} \int \dfrac{\sin (x+a-x)}{\sin x. \sin (x+a)} dx$$

$$\displaystyle =\dfrac{1}{ \sin a} \int \dfrac{ \sin (x+a) \cos x- \sin x\cos (x+a)}{\sin x . \sin (x+a)}dx$$

$$\displaystyle =\dfrac{1}{\sin a} \int [\cot x dx- \cot (x+a)] dx$$

$$=\dfrac{1}{\sin a} [\log |\sin x|-\log |\sin (x+a)| ] +c$$
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Subjective Medium Published on 17th 09, 2020
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