How can I quickly perform substitutions in integration?

Question

If i want to integrate
$$\int \sin x \cos x \, dx $$ and I let $u = \sin x$, then how do I work out what $dx$ is? I know that you can do \begin{align} \frac{du}{dx} &= \cos x \\[5pt] \frac{dx}{du} &= \frac{1}{\cos x} \\[5pt] dx &= \frac{du}{\cos x} \, , \end{align} but is there a shorter way that I can get there? I am asking because sometimes when I watch Youtube videos on integration by substitution where they can do it directly. Are they doing it mentally?

Answer 1

If $\frac{du}{dx}=\cos x$, then $du=\cos x \, dx$. Since the term $\cos x \, dx$ already appears in the integrand, there is no need to make $dx$ the subject of the equation. With practice, you can skip the $\frac{du}{dx}=\cos x$ step and simply write $du=\cos x \, dx$, meaning that you can perform the substitution mentally.

In this particular case, there are two viable alternatives to this method: as imranfat mentions in the comments, you can use the identity $\sin 2x=2\sin x \cos x$. Then, $$ \int\sin x \cos x \, dx = \int\frac{1}{2}\sin2x \, dx $$ At this stage, you could make the substitution $u=2x$, but that strikes me as a waste of time. Just note that $$ \frac{d}{dx}\left(-\cos2x\right)=2\sin2x $$ and so $$ \frac{d}{dx}\left(-\color{#F01C2C}{\frac{1}{4}}\cos2x\right)=\color{#F01C2C}{\frac{1}{2}}\sin2x $$ and you're finished. Again, with practice you can do this mentally.

The second method is to note that $$ \frac{d}{dx}\left(\sin^2x\right)=2\sin x\cos x \, , $$ from which we get $$ \int \sin x\cos x \, dx = \frac{\sin^2 x}{2}+C \, . $$ Note that $\frac{\sin^2 x}{2}$ and $-\frac{1}{4}\cos2x$ simply differ by a constant, meaning that they are both antiderivatives of the function in question.