Integrate the function sin (ax+b) cos (ax+b). | vedclass

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Question

(A) The given function is $I = \int \sin (ax+b) \cos (ax+b) \, dx$. We know that $\sin 2	heta = 2 \sin 	heta \cos 	heta$,so $\sin 	heta \cos 	het

Vedclass · Accepted Answer

$-\frac{1}{4a} \cos 2(ax+b) + C$

Vedclass · Answer

(A) The given function is $I = \int \sin (ax+b) \cos (ax+b) \, dx$. We know that $\sin 2	heta = 2 \sin 	heta \cos 	heta$,so $\sin 	heta \cos 	heta = \frac{\sin 2	heta}{2}$. Substituting $	heta = ax+b$,we get: $I = \int \frac{\sin 2(ax+b)}{2} \, dx = \frac{1}{2} \int \sin(2ax + 2b) \, dx$. Using the integration formula $\int \sin(kx+c) \, dx = -\frac{\cos(kx+c)}{k} + C$: $I = \frac{1}{2} \left( -\frac{\cos(2ax+2b)}{2a} ight) + C$. $I = -\frac{\cos 2(ax+b)}{4a} + C$,where $C$ is an arbitrary constant.

Integrate the function $\sin (ax+b) \cos (ax+b)$.

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