int sqrt{sin x} cos x , dx = frac{2}{3}(sin x | vedclass

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Question

(C) Given the integral is $I = \int \sqrt{\sin x} \cos x \, dx$. Let $t = \sin x$, then $dt = \cos x \, dx$. The integral becomes $I = \int \sqrt{t} \

Vedclass · Accepted Answer

$(2n\pi, (2n+1)\pi), n \in \mathbb{Z}$

Vedclass · Answer

(C) Given the integral is $I = \int \sqrt{\sin x} \cos x \, dx$. Let $t = \sin x$, then $dt = \cos x \, dx$. The integral becomes $I = \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C = \frac{2}{3} (\sin x)^{3/2} + C$. For the expression $\sqrt{\sin x}$ to be defined in the real number system, we must have $\sin x \ge 0$. Since the expression is in the denominator or part of a derivative chain where $\sin x$ must be strictly positive for the square root function to be differentiable, we consider $\sin x > 0$. The sine function $\sin x$ is positive in the first and second quadrants, which corresponds to the interval $(2n\pi, (2n+1)\pi)$ for any integer $n$.

$\int \sqrt{\sin x} \cos x \, dx = \frac{2}{3}(\sin x)^{3/2} + C$ is valid when $x$ lies in the interval

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