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

Option 1)
$(-sin\alpha ,cos\alpha )\;$

Option 2)
$\; (cos\alpha , sin\alpha )\;$

Option 3)
$\; (sin\alpha , cos\alpha )\;$

Option 4)
$\; (-cos\alpha ,sin\alpha )$

Answers (1)

best_answer

As learnt in concept

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

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

Put $x-\alpha =t$

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

= $sin\alpha \int cot\: t\:dt+cos\alpha \int dt$

= $sin\alpha ln\left | sint \right |+cos\alpha .(x-\alpha )+C$

Thus, $A=cos\alpha , B=sin\alpha$

Option 1)

$(-sin\alpha ,cos\alpha )\;$

This option is incorrect

Option 2)

$\; (cos\alpha , sin\alpha )\;$

This option is correct

Option 3)

$\; (sin\alpha , cos\alpha )\;$

This option is incorrect

Option 4)

$\; (-cos\alpha ,sin\alpha )$

This option is incorrect

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Aadil

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