Mathematics

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


ANSWER

$$\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sin x}}{{\sin \left( {x + \alpha } \right)}}} \right| + c$$


SOLUTION
Given $$I= \displaystyle\int \dfrac{1}{sin x. sin(x+\alpha )}dx$$

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

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

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

$$=\dfrac{1}{sin\alpha }[log|sinx|-log |sin (x+\alpha )|]+c$$

$$= cosec \, \alpha \, log \left|\dfrac{sin x}{sin (x+\alpha )}\right|+c$$
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Single Correct Medium Published on 17th 09, 2020
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