int x^2 sin x cos x , dx = | vedclass

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(C) We know that $\sin x \cos x = \frac{1}{2} \sin 2x$. Substituting this into the integral,we get: $I = \int x^2 \left(\frac{1}{2} \sin 2x\right) dx

Vedclass · Accepted Answer

$\frac{1-2x^2}{8} \cos 2x + \frac{x}{4} \sin 2x + c$

Vedclass · Answer

(C) We know that $\sin x \cos x = \frac{1}{2} \sin 2x$. Substituting this into the integral,we get: $I = \int x^2 \left(\frac{1}{2} \sin 2x\right) dx = \frac{1}{2} \int x^2 \sin 2x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,let $u = x^2$ and $dv = \sin 2x \, dx$. Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$. $I = \frac{1}{2} \left[ -\frac{x^2 \cos 2x}{2} - \int \left(-\frac{\cos 2x}{2}\right) (2x) \, dx \right]$ $I = -\frac{x^2 \cos 2x}{4} + \frac{1}{2} \int x \cos 2x \, dx$. Applying integration by parts again for $\int x \cos 2x \, dx$: Let $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$. $\int x \cos 2x \, dx = \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}$. Substituting this back: $I = -\frac{x^2 \cos 2x}{4} + \frac{1}{2} \left( \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right) + c$ $I = -\frac{x^2 \cos 2x}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + c$. Comparing this with the options,the expression can be rewritten as $\frac{1-2x^2}{8} \cos 2x + \frac{x}{4} \sin 2x + c$.

$\int x^2 \sin x \cos x \, dx =$

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