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.