evaluate the integral.\n int 4sinsqrt{x}dx \n int 4sinsqrt{x}dx=square

Explanation:

Step1: Use substitution

Let $u = \sqrt{x}$, then $x = u^{2}$ and $dx = 2u\ du$. The integral $\int 4\sin\sqrt{x}dx$ becomes $\int 4\sin(u)\cdot 2u\ du=8\int u\sin(u)du$.

Step2: Apply integration - by - parts

The integration - by - parts formula is $\int vdw=vw-\int wdv$. Let $v = u$ and $dw=\sin(u)du$. Then $dv = du$ and $w=-\cos(u)$. So, $8\int u\sin(u)du=8\left[-u\cos(u)+\int\cos(u)du\right]$.

Step3: Integrate $\cos(u)$

We know that $\int\cos(u)du=\sin(u)+C$. So, $8\left[-u\cos(u)+\int\cos(u)du\right]=8(-u\cos(u)+\sin(u))+C$.

Step4: Substitute back $u = \sqrt{x}$

We get $8(-\sqrt{x}\cos(\sqrt{x})+\sin(\sqrt{x}))+C$.

Answer:

$8(-\sqrt{x}\cos(\sqrt{x})+\sin(\sqrt{x}))+C$