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a
xsina+cosalogsinx−a+C
b
cosx+sinxlogsinx−a+C
c
(x−a)cosa+sinalogsin(x−a)+C
d
xcosa+sinalogsin(x−a)+C
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Correct option is D
I=∫sinxsin(x−a)dx=∫sin(x−a+a)sin(x−a)dx=∫sin(x−a)cosa+cos(x−a)sinasin(x−a)dx=∫[cosa+sinacot(x−a)]dx=cosa∫dx+sina∫cot(x−a)dx=xcosa+sinalog|sin(x−a)|+CSimilar Questions
The value of is equal to
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