Integrate the function sqrt{sin 2x} cos 2x. | vedclass

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Question

(A) Let $I = \int \sqrt{\sin 2x} \cos 2x \, dx$.Substitute $\sin 2x = t$.Differentiating both sides with respect to $x$,we get:$2 \cos 2x \, dx = dt$$

Vedclass · Accepted Answer

$\frac{1}{3}(\sin 2x)^{\frac{3}{2}} + C$

Vedclass · Answer

(A) Let $I = \int \sqrt{\sin 2x} \cos 2x \, dx$.Substitute $\sin 2x = t$.Differentiating both sides with respect to $x$,we get:$2 \cos 2x \, dx = dt$$\cos 2x \, dx = \frac{1}{2} dt$.Substituting these into the integral:$I = \int \sqrt{t} \cdot \frac{1}{2} dt$$I = \frac{1}{2} \int t^{\frac{1}{2}} dt$Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:$I = \frac{1}{2} \left( \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} \right) + C$$I = \frac{1}{2} \left( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right) + C$$I = \frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} + C$$I = \frac{1}{3} t^{\frac{3}{2}} + C$Substituting $t = \sin 2x$ back:$I = \frac{1}{3} (\sin 2x)^{\frac{3}{2}} + C$,where $C$ is the constant of integration.

Integrate the function $\sqrt{\sin 2x} \cos 2x$.

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