Find the volume of solid generated when $y=2 \sin(x^2)$ is rotated about the $x$-axis Find the volume of solid generated when $y=2 \sin(x^2)$ is rotated about the $x$ axis. The function intersects the $x$-axis at 0 and $\sqrt\pi$.
$$ \text{ Volume } = \int_0^\sqrt\pi \pi(2 \sin x^2)^2 dx=\int_0^\sqrt\pi 4\pi( \sin x^2)^2dx.$$
However I cannot figure out how to evaluate this integral. I tried substitution method and integration by parts with no luck. Any help is much appreciated.
 A: $4π\int sin²(x²) dx$
$2π\int 1dx - 2π\int 1-2sin²(x²)dx$
$2πx-2π\int cos(2x²)dx$
$u=\sqrt{2}x$
$du=\sqrt{2}dx$
$2πx-\sqrt{2}π\int cos(u²)du$
$2πx-\sqrt{2}πC(u)+c$ where C is the Cosine Fresnel Integral
$2πx-\sqrt{2}πC(\sqrt{2}x)+c$
$[2πx-\sqrt{2}πC(\sqrt{2}x)]_0^\sqrt{π}$
$2π\sqrt{π}-\sqrt{2}πC(\sqrt{2π})$
if it’d work better, there’s also a method you can use where instead of using fresnel integrals, you just add up error functions with complex arguments.  Comment below if you’d like me to show you.
Hope this helps 
EDIT:  the person who asked the question requested that I use the aforementioned error dunction method.  Let’s go back to this step:
$2πx-\sqrt{2}π\int cos(u²)du$
$cos(ø)=\frac{e^{øi}+e^{-øi}}2$
$2πx-\sqrt{2}π\int \frac{e^{u²i}+e^{-u²i}}2 du$
$2πx-\frac{π}{\sqrt{2}}\int e^{u²i}du-\frac{π}{\sqrt{2}}\int e^{-u²i}du$
$2πx-\sum \frac{π}{\sqrt{2}}\int e^{±u²i}du$
$v=\frac{(1∓i)u}{\sqrt{2}}$
$u=\frac{(1±i)v}{\sqrt{2}}$
$du=\frac{1±i}{\sqrt{2}}dv$
$2πx-\sum \frac{π}{\sqrt{2}}\int e^{±(\frac{(1±i)v}{\sqrt{2}})²i}•\frac{1±i}{\sqrt{2}}dv$
$2πx-\sum \frac{π(1±i)}{2}\int e^{-v²}dv$
$\int e^{-v²}dv=\frac{2erf(v)}{\sqrt{π}}$
$2πx-\sum \frac{π(1±i)}{2}•\frac{2erf(v)}{\sqrt{π}}$
$2πx-\sum \sqrt{π}(1±i)erf(\frac{(1∓i)u}{\sqrt{2}})$
$2πx-\sqrt{π}(1+i)erf(\frac{(1-i)u}{\sqrt{2}})-\sqrt{π}(1-i)erf(\frac{(1+i)u}{\sqrt{2}})$
$u=\sqrt{2}x$
$2πx-\sqrt{π}(1+i)erf((1-i)x)-\sqrt{π}(1-i)erf((1+i)x)+C$
