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?
Best Answer
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!