Question

$int\frac{cos^{3}sqrt{t}sinsqrt{t}}{sqrt{t}}dt$

Explanation:

Step1: Use substitution

Let $u = \sqrt{t}$, then $t = u^{2}$ and $dt = 2u\ du$. The integral becomes $\int\frac{\cos^{3}u\sin u}{u}\cdot2u\ du=2\int\cos^{3}u\sin u\ du$.

Step2: Use another - substitution

Let $v=\cos u$, then $dv = -\sin u\ du$. The integral $2\int\cos^{3}u\sin u\ du=- 2\int v^{3}dv$.

Step3: Integrate

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $-2\int v^{3}dv=-2\times\frac{v^{4}}{4}+C=-\frac{1}{2}v^{4}+C$.

Step4: Back - substitute

Substitute back $v = \cos u$ and $u=\sqrt{t}$, we get $-\frac{1}{2}\cos^{4}\sqrt{t}+C$.

Answer:

$-\frac{1}{2}\cos^{4}\sqrt{t}+C$