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Can someone be kind enough to show me the steps to integrate this, I'm sure it's by parts but how do I split up the sin part? $$x\sin(1+2x)$$

Aapeli
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Helpthanks
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    I would try this: let $u=x$ and $v'=\sin(1+2x)$, then integrate by parts ($\int{uv'dx}=uv-\int{u'vdx}$). – Aapeli Jun 19 '14 at 09:39
  • @Helpthanks If you found an answer helpful then you should upvote it, so that others can see that you found it helpful. You can also "accept" an answer if you feel that it answers your question sufficiently. – user1729 Aug 05 '14 at 09:41

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Let $u=x$ and $dv =\sin (1+2x) \, dx$ Then $v=-\frac{1}{2}\cos(1+2x)$. Now perform integration by parts.

Anurag A
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Let $u=x$ and $v'=\sin(1+2x)$, then $u'=2$, $v=-\frac{\cos(1+2x)}{2}$: \begin{align}\int{uv'\ dx}&=uv-\int{u'v\ dx}\\ \int{x\sin(1+2x)\ dx}&=-\frac{x\cos(1+2x)}{2}-\int{-\frac{2\cos(1+2x)}{2}\ dx}+C\\ &=-\frac{x\cos(1+2x)}{2}+\frac{\sin(1+2x)}{4}+C\\ \therefore\int{x\sin(1+2x)\ dx}&=\frac{1}{4}\sin(1+2x)-\frac{1}{2}x\cos(1+2x)+C \end{align}

Aapeli
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I would start with an integration by substitution first. So, let $z=1+2x$ and note that this implies $x=\frac{z-1}{2}$, so we obtain the following. $$\frac14\int(z-1)\sin z\operatorname{d}z=\frac14\int z\sin\operatorname{d}z-\frac14\int \sin z\operatorname{d}z$$ Then, integrating $\sin z$ is standard, while $z\sin z$ is a standard easy example of integration by parts (and is probably in your notes!).

user1729
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  • If I'm evaluating a double integral in Cartesians, would I evaluate the trig in degrees or radians? – Helpthanks Jun 19 '14 at 11:33
  • Theoretically, there is no difference. However, certain formulae may require an input in radians. So it depends on what you are going to do with your answer. – user1729 Jun 19 '14 at 12:49