evaluate $intsqrt{sin x}cos x dx$

Explanation:

Step1: Use substitution

Let $u = \sin x$, then $du=\cos xdx$.

Step2: Rewrite the integral

The integral $\int\sqrt{\sin x}\cos xdx$ becomes $\int\sqrt{u}du$.

Step3: Integrate $\sqrt{u}$

We know that $\sqrt{u}=u^{\frac{1}{2}}$, and $\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+ 1}}{\frac{1}{2}+1}+C=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{3}u^{\frac{3}{2}}+C$.

Step4: Substitute back $u = \sin x$

We get $\frac{2}{3}(\sin x)^{\frac{3}{2}}+C$.

Answer:

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