$\int x^{2} \sin(x^{2} +2) \mathrm{d}x$

Question

$$\int x^{2} \sin(x^{2} +2) \mathrm{d}x$$

so, i think i have to do $u$ sub on the equation in $\sin$, but from there i'm not sure how to approach integration by parts afterwards...

answer might be wrong but i got $$u = x^{2} +2$$ $$\mathrm{d}u = 2x$$ $$\mathrm{d}x = \frac{\mathrm{d}u}{2x}$$ $$x^{2} = u-2$$ i plug into the original equation $$\frac{1}{2} \int u\sin(u) \mathrm{d}u$$ $$(u-2) \sin(u-2)$$

...yeah i know its wrong just not sure how to approach this problem

Answer 1

The problem is much more complex than it looks (almost as Aaron M suggested).

Considering $$I=\int x^2 \sin \left(x^2+2\right)\,dx=\int x \sin \left(x^2+2\right) x \,dx$$ So, integrating by parts $$u=x \qquad du=dx \qquad dv=\sin \left(x^2+2\right) x \,dx\qquad v=-\frac 12 \cos(x^2+2)$$ make $$I=-\frac x2 \cos(x^2+2)+\frac 12 \int \cos(x^2+2) \,dx$$ Now expand $$\cos(x^2+2)=\cos (2) \cos \left(x^2\right)-\sin (2) \sin \left(x^2\right)$$ and we face the problem of computing $$J=\int \cos \left(x^2\right)\,dx \qquad , \qquad K=\int \sin \left(x^2\right)\,dx$$ which are not simple $$J=\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} x\right)\qquad , \qquad K=\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} x\right)$$ where appear the cosine and sine Fresnel integrals.

Combining all of that make leads to $$I=-\frac x2 \cos(x^2+2)+\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\cos (2)\, C\left(\sqrt{\frac{2}{\pi }} x\right)-\sin (2)\, S\left(\sqrt{\frac{2}{\pi }} x\right)\right)+C$$

There is probably a typo and the problem should probably be instead $$I=\int x \sin \left(x^2+2\right)\,dx$$ which is quite simple using $$I=\frac 12\int \sin(x^2+2)\, d(x^2)=\frac 12\int \sin(y+2)\, dy=\frac 12\int \sin(z)\, dz=-\frac 12 \cos(z)+C$$ What is amazing is that $$I_n=\int x^{2n+1}\sin(x^2+2)\,dx$$ would not present any major problem $$I_1=\frac{1}{2} \left(\sin \left(x^2+2\right)-x^2 \cos \left(x^2+2\right)\right)$$ $$I_2=x^2 \sin \left(x^2+2\right)-\frac{1}{2} \left(x^4-2\right) \cos \left(x^2+2\right)$$ $$I_3=\frac{1}{2} \left(3 \left(x^4-2\right) \sin \left(x^2+2\right)-x^2 \left(x^4-6\right) \cos \left(x^2+2\right)\right)$$ while $$J_n=\int x^{2n}\sin(x^2+2)\,dx$$ always involves Fresnel integrals.

Answer 2

First get rid of the constant term $2$ by means of the addition formula and solve the integral for $\cos$ and $\sin$.

Then by parts,

$$\int \frac x2(2x\sin x^2)\, dx=-\frac x2\cos x^2+\frac12\int\cos x^2\,dx$$ and similarly for the sine.

The final integrals are the well-known Fresnel functions which have no closed form, and there is nothing more that you can do. (https://en.wikipedia.org/wiki/Fresnel_integral)