int_0^{pi /3} cos 3x , dx = | vedclass

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(A) To evaluate the definite integral $\int_0^{\pi /3} \cos 3x \, dx$,we first find the antiderivative of $\cos 3x$.The integral of $\cos 3x$ is $\fra

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) To evaluate the definite integral $\int_0^{\pi /3} \cos 3x \, dx$,we first find the antiderivative of $\cos 3x$.The integral of $\cos 3x$ is $\frac{\sin 3x}{3}$.Applying the fundamental theorem of calculus:$\int_0^{\pi /3} \cos 3x \, dx = \left[ \frac{\sin 3x}{3} ight]_0^{\pi /3}$$= \frac{\sin(3 	imes \frac{\pi}{3})}{3} - \frac{\sin(3 	imes 0)}{3}$$= \frac{\sin(\pi)}{3} - \frac{\sin(0)}{3}$$= \frac{0}{3} - \frac{0}{3} = 0$.Thus,the correct option is $A$.

$\int_0^{\pi /3} \cos 3x \, dx = $

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