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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
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