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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.)

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  • $\begingroup$ Do a substitution instead. Let $u=x^4$. $\endgroup$ – André Nicolas Jun 22 '14 at 13:59
  • $\begingroup$ thanks a lot, you guys are really good. comparetively at least. $\endgroup$ – qwertz Jun 22 '14 at 14:12
  • $\begingroup$ You are welcome. I have noticed in teaching calculus that after seeing integration by parts, some students really like the technique, and try to use it even when it is not suitable, or not optimal. The thing to notice is that the derivative of the "ugly" $x^4$ is basically sitting in front. That's a signal that substitution might be useful. $\endgroup$ – André Nicolas Jun 22 '14 at 14:18
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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 $$\int \sin(\underbrace{x^4}_{\large = u})\underbrace{\left(x^3\,dx\right)}_{\large = \frac {du}{4}} = \frac 14\int \sin u \,du$$

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

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HINT:

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

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HINT:

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

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

Hint: note that

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

so...

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