Answer
$\int S(x) dx=S(x)\cdot x+\frac{1}{\pi}\cos (\frac{1}{2}\pi x^2)+C$
Work Step by Step
$S(x)=\int_0^x\sin(\frac{1}{2}\pi t^2)dt$
Using the Fundamental Theorem of Calculus in Part 1, we have
$\frac{dS}{dx}=\sin(\frac{1}{2}\pi x^2)$
$dS=\sin(\frac{1}{2}\pi x^2)dx$
Now, evaluate the given integral using the Integration by Parts:
$\int S(x) dx=S(x)\cdot x-\int x dS$
$\int S(x) dx=S(x)\cdot x-\int x \cdot \sin (\frac{1}{2}\pi x^2) dx$ (Let $u=\frac{1}{2}\pi x^2$. Then, $du=\pi x dx$ or $dx =\frac{du}{\pi x}$)
$\int S(x) dx=S(x)\cdot x-\int x \cdot \sin (u) \cdot \frac{du}{\pi x}$
$\int S(x) dx=S(x)\cdot x-\int \frac{1}{\pi} \sin (u)du$
$\int S(x) dx=S(x)\cdot x+\frac{1}{\pi}\cos u+C$
$\int S(x) dx=S(x)\cdot x+\frac{1}{\pi}\cos (\frac{1}{2}\pi x^2)+C$
