evaluate $intsqrt{cos x}sin^{3}x dx$.

Explanation:

Step1: Rewrite $\sin^{3}x$

We know that $\sin^{3}x=\sin x\cdot\sin^{2}x=\sin x(1 - \cos^{2}x)$. So the integral becomes $\int\sqrt{\cos x}\sin x(1 - \cos^{2}x)dx$.

Step2: Use substitution

Let $u = \cos x$, then $du=-\sin xdx$. The integral $\int\sqrt{\cos x}\sin x(1 - \cos^{2}x)dx$ can be rewritten as $-\int\sqrt{u}(1 - u^{2})du=-\int(u^{\frac{1}{2}}-u^{\frac{5}{2}})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: [ \begin{align*} -\int(u^{\frac{1}{2}}-u^{\frac{5}{2}})du&=-\left(\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{u^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right)+C\ &=-\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}-\frac{u^{\frac{7}{2}}}{\frac{7}{2}}\right)+C\ &=-\frac{2}{3}u^{\frac{3}{2}}+\frac{2}{7}u^{\frac{7}{2}}+C \end{align*} ]

Step4: Substitute back $u=\cos x$

The result is $-\frac{2}{3}\cos^{\frac{3}{2}}x+\frac{2}{7}\cos^{\frac{7}{2}}x + C$.

Answer:

$-\frac{2}{3}\cos^{\frac{3}{2}}x+\frac{2}{7}\cos^{\frac{7}{2}}x + C$