The only way I could think of to integrate $xe^{\sin x}\cos x$ with respect to $x$ was by integration by parts, by substituting $$\sin x=t$$ so that $$\cos x\ \mathrm{d}x=\mathrm{d}t$$ so that it works out to $$\int xe^t\ \mathrm{d}t$$ and by taking $e^t\ \mathrm{d}t$ as $\mathrm{d}v$ in the following general equation for integration by parts.

$$\int u\ \mathrm{d}v= u\int \mathrm{d}v - \int v\ \mathrm{d}u$$ This sort of doesn't make sense, even if it seems to work okay; with respect to what variable do you differentiate $x$ to get $1$ in the second term(in the compact equation, this refers to calculating the $\mathrm{d}u$ term)? I did it with respect to $\mathrm{d}x$, but there wasn't a $\mathrm{d}x$ there, so it seems wrong. Moreover, after getting $u$ by integrating $\mathrm{d}u$, and taking the derivative of $x$ to get $1$, you have $\displaystyle\int e^t$ without a $\mathrm{d}x$.

Where did I mess up here, and how do I integrate that?

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    $\begingroup$ it should be $\int u dv = u v - \int v du$ $\endgroup$ – janmarqz Apr 4 at 3:45
  • $\begingroup$ Fixed that, I saw the error, thanks. $\endgroup$ – harry Apr 4 at 3:46
  • $\begingroup$ $\int uvdx=u\int vdx-\int \frac{du}{dx}\left(\int vdx\right) dx$ - isn't this the correct formua for integration by parts? $\endgroup$ – Ishraaq Parvez Apr 4 at 3:46
  • $\begingroup$ @Ishraaq Parvez. No that is a different rule. $\endgroup$ – Rounak Sarkar Apr 4 at 3:48
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    $\begingroup$ Wait, in the third line where you've written $\int xe^tdt$, won't that be $\int\arcsin te^tdt$? You've substituted $\sin x=t\implies x=\arcsin t$ $\endgroup$ – Debartha Paul Apr 4 at 4:09

Based on your attempt, we can integrate by parts the proposed function to obtain: \begin{align*} \int x\exp(\sin(x))\cos(x)\mathrm{d}x = x\exp(\sin(x)) - \int\exp(\sin(x))\mathrm{d}x \end{align*}

Unfortunately, the last integral cannot be solved in terms of elementary functions.

Hopefully this helps!

  • $\begingroup$ why don't you upvote the Q? $\endgroup$ – janmarqz Apr 4 at 4:00

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