int {2sin x cos x} ,dx is equal to | vedclass

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(D) We know that $2\sin x \cos x = \sin 2x$. Substituting this into the integral,we get: $I = \int \sin 2x \,dx$ Using the integration formula $\int \

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$\sin^2 x + c$

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(D) We know that $2\sin x \cos x = \sin 2x$. Substituting this into the integral,we get: $I = \int \sin 2x \,dx$ Using the integration formula $\int \sin(ax) \,dx = -\frac{\cos(ax)}{a} + c$,we have: $I = -\frac{\cos 2x}{2} + c$ Alternatively,using the identity $\cos 2x = 1 - 2\sin^2 x$: $I = -\frac{1 - 2\sin^2 x}{2} + c$ $I = -\frac{1}{2} + \sin^2 x + c$ Since $c$ is an arbitrary constant,$- \frac{1}{2} + c$ can be written as a new constant $C$. Thus,$I = \sin^2 x + C$. Therefore,the correct option is $(d)$.

$\int {2\sin x \cos x} \,dx$ is equal to

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