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Question

Question: \[\int_{}^{}{\sin x\sin hxdx =}\]...

∫sin⁡xsin⁡hxdx=\int_{}^{}{\sin x\sin hxdx =}

A

12(ex−e−x)+c\frac{1}{2}(e^{x} - e^{- x}) + c

B

12[sin⁡xcos⁡hx−cos⁡xsin⁡hx]+c\frac{1}{2}\lbrack\sin x\cos hx - \cos x\sin hx\rbrack + c

C

12[sin⁡xcos⁡hx+cos⁡xcos⁡hx]+c\frac{1}{2}\lbrack\sin x\cos hx + \cos x\cos hx\rbrack + c

D

None of these

Answer

12[sin⁡xcos⁡hx−cos⁡xsin⁡hx]+c\frac{1}{2}\lbrack\sin x\cos hx - \cos x\sin hx\rbrack + c

Explanation

Solution

I=∫sin⁡xsin⁡hxdx=−sin⁡hxcos⁡x+∫cos⁡hxcos⁡xdx=sin⁡hxcos⁡x+[cos⁡hxsin⁡x−∫sin⁡hxsin⁡xdx]I=12[sin⁡xcos⁡x−cos⁡xsin⁡hx]+c.I = \int_{}^{}{\sin x\sin hxdx = - \sin hx\cos x} + \int_{}^{}\begin{aligned} & \cos hx\cos xdx = \sin hx\cos x + \\ & \lbrack\cos hx\sin x - \int_{}^{}{\sin hx\sin xdx\rbrack} \end{aligned}I = \frac{1}{2}\lbrack\sin x\cos x - \cos x\sin hx\rbrack + c.