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
Questions 203525
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