We will evaluate $\displaystyle\int
3x^2\sqrt{x^3-1}\,dx$. We choose $u$ with the idea that
the derivative of $u$ is part of the integrand. Here, it
seems that $u=x^3-1$ might
work. We try it, and after substitution (check that we substituted exactly and
correctly), we have $$\int
3x^2\sqrt{x^3-1}\,dx\overset{\fbox{$ \,\,u\,=\,x^3-1\\
du\,=\,3x^2\,dx$}\\}{=} \int\sqrt u\,du=\int
u^{\frac{1}{2}}\,du=\frac{2}{3}u^{\frac{3}{2}}+c=\frac{2}{3}(x^3-1)^{\frac{3}{2}}+c.$$
Now check:
$\frac{d}{dx}\left(\frac{2}{3}(x^3-1)^{\frac{3}{2}}+c\right)=\frac{2}{3}\frac{3}{2}(x^3-1)^{\frac{1}{2}}(3x^2),$
which is our original integrand. **You
may want to make it a habit to check your antiderivatives,
thereby avoiding errors.**

** **

**DO:** Now you try to
find $\int 2(\cos x) e^{\sin x}\,dx$. What might
$u$ be? The rest of the integrand must be **multiplied** by $du$, so **$du$ can't be an exponent, an argument of
a function, or a denominator**. Do not look
ahead until you have tried. Check your answer!

$\int
2(\cos x )e^{\sin x}\,dx\overset{\fbox{$ \,\,\,u\,=\,\sin x\\
\,du\,=\,\cos x\,dx\\2du\,=\,2\cos x\,dx$}\\}{=}\int e^u\cdot
2\,du=2e^u+c=2e^{\sin x}+c$.

Notice we multiplied by 2 in our substitution to
get $2du$ since $2\cos x\,dx$ is in our integrand.

Checking our answer:
$\frac{d}{dx}(2e^{\sin x}+c)=2e^{\sin x}\cos x$. Yay!

**Fact: ** **We can multiply or divide both sides of our
substitution by any constant. **

**DO:** One more.
Find $\int x\cos(6x^2)\,dx.$ Do not look ahead until
you've tried.

$\int x\cos(6x^2)\,dx\overset{\fbox{$ \quad u\,=\,6x^2\\
\,\,\,\,
du\,=\,12x\,dx\\\frac{1}{12}du\,=\,x\,dx$}\\}{=}\int\frac{1}{12}\cos
u\,du=\frac{1}{12}\sin u+c=\frac{1}{12}\sin(6x^2)+c$

We divided both sides by 12 to get the left-hand side of our
substitution (with $du$) to match the integrand.

Check:
$\frac{d}{dx}\left(\frac{1}{12}\sin(6x^2)+c\right)=\frac{1}{12}\cos(6x^2)(12x)$.
Yay!

**Our goal when substituting is to
get the exact integrand. **