Find int cos 6x sqrt{1+sin 6x} , dx | vedclass

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Question

(A) Let $t = 1 + \sin 6x$. Then,differentiating with respect to $x$,we get $dt = 6 \cos 6x \, dx$,which implies $\cos 6x \, dx = \frac{1}{6} \, dt$. S

Vedclass · Accepted Answer

$\frac{1}{9}(1+\sin 6x)^{\frac{3}{2}}+C$

Vedclass · Answer

(A) Let $t = 1 + \sin 6x$. Then,differentiating with respect to $x$,we get $dt = 6 \cos 6x \, dx$,which implies $\cos 6x \, dx = \frac{1}{6} \, dt$. Substituting these into the integral: $\int \cos 6x \sqrt{1+\sin 6x} \, dx = \int \sqrt{t} \cdot \frac{1}{6} \, dt$ $= \frac{1}{6} \int t^{\frac{1}{2}} \, dt$ $= \frac{1}{6} \left( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right) + C$ $= \frac{1}{6} \cdot \frac{2}{3} t^{\frac{3}{2}} + C$ $= \frac{1}{9} (1 + \sin 6x)^{\frac{3}{2}} + C$

Find $\int \cos 6x \sqrt{1+\sin 6x} \, dx$

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