# Integration by parts- integrating without a $\mathrm{d}x$?

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?

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