Stuck on an integral $4x\sin (x^2)\cos (x^2)$

Question

I'm trying to get my head around this integral but it just doesn't click $$4x\sin (x^2)\cos (x^2) $$

I have tried substitution but I am confusing myself!

Do I substitute $\cos (x^2)$ getting $du=- 2x\sin (x^2) $ but I'm not sure where to go from there.

Answer 1

The substitution $x^2=t$ gives $$-\cos^2 x^2+C$$.

Answer 2

Let $$ u = x^{2} \qquad \Rightarrow \qquad du = 2x dx $$ Then the primitive becomes $$ \int 4 x \sin \left(x^2\right) \cos \left(x^2\right) \, dx \Rightarrow \int 2 \sin u \cos u \, du = -\frac{1}{2} \cos (2 u) \, du \Rightarrow -\frac{1}{2} \cos \left(2 x^2\right) $$

Answer 3

$4x\sin(x^{2})\cos(x^{2})=2x\sin(2x^{2})$. Now you can take the substitution $y=x^{2}$.

Answer 4

$$\int 4x\sin (x^2)\cos (x^2) \,dx $$ $$= \int 2\sin (x^2) \cos(x^2) \, dx^2$$ $$=\int 2\sin (x^2) \, d\sin(x^2) $$ $$= \sin^2(x^2) + C $$