int sqrt{2 + sin 3x} cdot cos 3x , dx = | vedclass

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Question

(D) Let $I = \int \sqrt{2 + \sin 3x} \cdot \cos 3x \, dx$. Substitute $t = 2 + \sin 3x$. Then,differentiating with respect to $x$,we get $dt = 3 \cos

Vedclass · Accepted Answer

$\frac{2}{9}(2 + \sin 3x)^{3/2} + c$

Vedclass · Answer

(D) Let $I = \int \sqrt{2 + \sin 3x} \cdot \cos 3x \, dx$. Substitute $t = 2 + \sin 3x$. Then,differentiating with respect to $x$,we get $dt = 3 \cos 3x \, dx$,which implies $\cos 3x \, dx = \frac{1}{3} dt$. Substituting these into the integral: $I = \int \sqrt{t} \cdot \frac{1}{3} dt = \frac{1}{3} \int t^{1/2} dt$. Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + c$: $I = \frac{1}{3} \cdot \frac{t^{3/2}}{3/2} + c = \frac{1}{3} \cdot \frac{2}{3} t^{3/2} + c = \frac{2}{9} t^{3/2} + c$. Substituting back $t = 2 + \sin 3x$: $I = \frac{2}{9}(2 + \sin 3x)^{3/2} + c$.

$\int \sqrt{2 + \sin 3x} \cdot \cos 3x \, dx = $

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