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)

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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