integration by parts of trig functions

Question

Can anyone help me with this integral?

$\int{x^3 \sin(x^4) dx}$

I set $u=x^3$, and I let $v=-\cos(x^4)$, so that $\frac{dv}{dx}=\sin(x^4)$

I tried using integration by parts, but, whenever I come to the term where I have to integrate $\frac{du}{dx}(v)$, I get $(\ldots) + \int{3x^2 \cos(x^4) dx}$, which means I'll have to use integration by parts again, and it'll be a never-ending spiral. (I could be wrong.)

Answer 1

HINT:

Set $\displaystyle x^4=y\implies 4x^3\ dx=dy$

Answer 2

HINT:

The integrand is of the form $[\frac{1}{4}f'(x)\sin(f(x))]$

Answer 3

Does differentiating $\cos{f(x)}$ tell you something?

Hint: note that

$$- \frac{\mathrm{d} \cos{f} }{\mathrm{d} x} = f' \, \sin{f}, $$

so...

Answer 4

Notice that $(x^4)' = 4x^3$.

So no need for integration by parts!

The simple u-substitution of $u = x^4 \implies du = 4(x^3)\,dx \iff x^3\,dx = \dfrac{du}{4}$.

That gives us the integral $$\int \sin(\underbrace{x^4}_{\large = u})\underbrace{\left(x^3\,dx\right)}_{\large = \frac {du}{4}} = \frac 14\int \sin u \,du$$

$$= -\frac 14\cos(u) + C = -\frac 14\cos(x^4) + C$$