MHT CET · Maths · Indefinite Integration
If \(\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+c\), then the value of \(\mathrm{A}\) and \(\mathrm{B}\) are respectively (where \(\mathrm{c}\) is a constant of integration)
- A \(\cos \alpha, \sin \alpha\)
- B \(\sin \alpha, \cos \alpha\)
- C \(-\cos \alpha, \sin \alpha\)
- D \(-\sin \alpha, \cos \alpha\)
Answer & Solution
Correct Answer
(A) \(\cos \alpha, \sin \alpha\)
Step-by-step Solution
Detailed explanation
\(\text {Let } I=\int \frac{\sin x}{\sin (x-\alpha)} d x \)
\( =\int \frac{\sin [(x-\alpha)+\alpha]}{\sin (x-\alpha)} d x=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \)
\( =\cos \alpha \int d x+\sin \alpha \int \frac{\cos (x-\alpha)}{\sin (x-\alpha)} d x=(\cos \alpha)(x)+(\sin \alpha) \)
\( \log |\sin (x-\alpha)|+c\)
Comparing with given data, we get \(\mathrm{A}=\cos \alpha\) and \(\mathrm{B}=\sin \alpha\)
\( =\int \frac{\sin [(x-\alpha)+\alpha]}{\sin (x-\alpha)} d x=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \)
\( =\cos \alpha \int d x+\sin \alpha \int \frac{\cos (x-\alpha)}{\sin (x-\alpha)} d x=(\cos \alpha)(x)+(\sin \alpha) \)
\( \log |\sin (x-\alpha)|+c\)
Comparing with given data, we get \(\mathrm{A}=\cos \alpha\) and \(\mathrm{B}=\sin \alpha\)
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