Mathematics

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


ANSWER

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


SOLUTION
let $$x-\alpha=t=>x=t-\alpha$$
and $$dx=dt$$

$$\int \dfrac{sin(t-\alpha)}{sint}dt$$

$$=>\int \dfrac{sintcos\alpha+costsin\alpha}{sint}dt$$

$$=>\int \dfrac{sintcos\alpha}{sint}dt+\int \dfrac{costsin\alpha}{sint}dt$$

$$=>cos\alpha \int dt+sin\alpha \int cost dt$$

$$=>t\times cos\alpha + sin\alpha \times log|sint|+c$$

now put the value of $$t$$

$$=>(x-\alpha)cos\alpha+sin\alpha \times log|sin(x-\alpha)|+c$$

hence,

$$=>a=cos\alpha$$

$$=>b=sin\alpha$$




























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Single Correct Medium Published on 17th 09, 2020
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