Mathematics

If $$\int {\frac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \sin \left( {x - \alpha } \right) + C} ,$$ then value of $$(A,B)$$ is


ANSWER

$$\left( {\cos \alpha ,\sin \alpha } \right)$$


SOLUTION

$$\\\int(\frac{sinx}{sin(x-\alpha)})\>dx$$

$$\\let (x-\alpha)=t$$

$$\\\therefore x=\alpha+t$$

$$\\dx=dt$$

$$\\I=\int(\frac{sin(\alpha+t)}{sin(t)})\>dx$$

$$\\=\int(\frac{sin\alpha cost+cos\alpha sint}{sint})\>dx$$

$$\\=\int sin\alpha (\frac{cost}{sint})\>dt+\int cos\alpha\>dt$$

$$\\=sin\alpha \int (\frac{cost}{sint})\>dt+cos\alpha \int \>dt$$

$$\\=sin\alpha \>log(sint)+cos\alpha \>(t)+C$$

$$\\=sin\alpha \>logsin(x-\alpha)+cos\alpha\>(x-\alpha)+C$$

$$\\=x\>cos\alpha+sin\alpha\>logsin(x-\alpha)+C_{1}$$

Compare it with $$Ax+B\>logsin(x-\alpha)+C$$

we get A=$$cos\alpha\>and\>B=sin\alpha$$

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Single Correct Medium Published on 17th 09, 2020
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