Finding Antiderivatives: The Fundamental Theorem reduces the problem of finding area under curves to finding antiderivatives.
Thus, techniques for finding antiderivatives (indefinite integrals) will play a central role in this course.
Today, we learn our first antiderivative technique.
Recall: The Chain Rule $\displaystyle \frac{d}{dx}[f(g(x))]=f'(g(x)) g'(x).$
Restating the above in integral form gives us...
The Reverse Chain Rule
$$\int f'(g(x)) g'(x) \, dx=f(g(x))+C$$
Restating this in a slightly different form gives us our first integration technique.
Integration by Substitution
Let $u = g(x)$ where $g'(x)$ is continuous over an interval, $f (x)$ be continuous over the corresponding range of $g$, and let $F(x)$ be an antiderivative of $f (x).$ Then, $$\int f(g(x)) g'(x) \, dx=\int f(u) \, du =F(u)+C=F(g(x))+C$$ Note: When we let $u=g(x),$ this is the substitution that gives the technique its name.
Example
Use a suitable change of variables to determine the indefinite integral $$\int 6x (3x^2+4)^4 \, dx$$ Note: The "outside" function $6x$ is the derivative of the "inside" function $3x^2+4.$
Let $u=3x^2+4.$ Then $\displaystyle \frac{du}{dx}=6x$ so that $du=6x\,dx.$
Thus,
$$
\begin{array}{lll}
\displaystyle \int 6x (3x^2+4)^4 \, dx&=\displaystyle \int (3x^2+4)^4 \cdot 6x\, dx&\mbox{}\\
&=\displaystyle \int u^4 du &\mbox{by the above substitution}\\
&=\displaystyle \frac{1}{5}u^5+C &\mbox{}\\
&=\displaystyle \frac{1}{5}(3x^2+4)^5+C &\mbox{by the above substitution}\\
\end{array}
$$
Reminder: You may always check your work by differentiation: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\frac{1}{5}(3x^2+4)^5+C\right) &\displaystyle=\frac{1}{5}\frac{d}{dx}(3x^2+4)^5+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{5}\cdot 5(3x^2+4)^4\frac{d}{dx}(3x^2+4)+0&\mbox{}\\ &\displaystyle=(3x^2+4)^4 6x&\mbox{}\\ \end{array} $$
Reminder: You may always check your work by differentiation: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\frac{1}{5}(3x^2+4)^5+C\right) &\displaystyle=\frac{1}{5}\frac{d}{dx}(3x^2+4)^5+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{5}\cdot 5(3x^2+4)^4\frac{d}{dx}(3x^2+4)+0&\mbox{}\\ &\displaystyle=(3x^2+4)^4 6x&\mbox{}\\ \end{array} $$
Example
Use a suitable change of variables to determine the indefinite integral $$\int \cos(x^3) x^2 \, dx$$ Note: The "outside" function $x^2$ is a constant multiple of the derivative of the "inside" function $x^3.$ This one of the dead giveaways that substitution will work.
Let $u=x^3.$ Then $\displaystyle \frac{du}{dx}=3x^2$ so that $\displaystyle \frac{1}{3}du=x^2\,dx.$
Thus,
$$
\begin{array}{lll}
\displaystyle \int \cos(x^3) x^2 \, dx&=\displaystyle \int \cos(u)\cdot \frac{1}{3}du&\mbox{}\\
&=\displaystyle \frac{1}{3}\int \cos(u)\, du&\mbox{}\\
&=\displaystyle \frac{1}{3}\sin(u)+C&\mbox{}\\
&=\displaystyle \frac{1}{3}\sin(x^3)+C&\mbox{}\\
\end{array}
$$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\frac{1}{3}\sin(x^3)+C\right) &\displaystyle=\frac{1}{3}\frac{d}{dx}\sin(x^3)+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{3}\cos(x^3)\frac{d}{dx}x^3+0&\mbox{}\\ &\displaystyle=\frac{1}{3}\cos(x^3)\cdot 3x^2&\mbox{}\\ &\displaystyle=\cos(x^3)x^2&\mbox{}\\ \end{array} $$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\frac{1}{3}\sin(x^3)+C\right) &\displaystyle=\frac{1}{3}\frac{d}{dx}\sin(x^3)+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{3}\cos(x^3)\frac{d}{dx}x^3+0&\mbox{}\\ &\displaystyle=\frac{1}{3}\cos(x^3)\cdot 3x^2&\mbox{}\\ &\displaystyle=\cos(x^3)x^2&\mbox{}\\ \end{array} $$
Example
Use a suitable change of variables to determine the indefinite integral $$\displaystyle \int (2x-3)^{-7} \, dx$$
Let $u=2x-3.$ Then $\displaystyle \frac{du}{dx}=2$ so that $\displaystyle \frac{1}{2}du=\,dx.$
Thus,
$$
\begin{array}{lll}
\displaystyle \int (2x-3)^{-7} \, dx&=\displaystyle \int u^{-7}\cdot \frac{1}{2}du&\mbox{}\\
&=\displaystyle \frac{1}{2}\int u^{-7}\, du&\mbox{}\\
&=\displaystyle \frac{1}{2}\left(-\frac{1}{6}u^{-6}\right)+C&\mbox{}\\
&=\displaystyle \frac{1}{2}\left(-\frac{1}{6}(2x-3)^{-6}\right)+C&\mbox{}\\
&=\displaystyle -\frac{1}{12}(2x-3)^{-6}+C&\mbox{}\\
\end{array}
$$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(-\frac{1}{12}(2x-3)^{-6}+C\right) &\displaystyle=-\frac{1}{12}\frac{d}{dx}(2x-3)^{-6}+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=-\frac{1}{12}(-6)(2x-3)^{-7}\frac{d}{dx}(2x-3)+0&\mbox{}\\ &\displaystyle=\frac{1}{2}(2x-3)^{-7}\cdot 2&\mbox{}\\ &\displaystyle=(2x-3)^{-7}&\mbox{}\\ \end{array} $$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(-\frac{1}{12}(2x-3)^{-6}+C\right) &\displaystyle=-\frac{1}{12}\frac{d}{dx}(2x-3)^{-6}+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=-\frac{1}{12}(-6)(2x-3)^{-7}\frac{d}{dx}(2x-3)+0&\mbox{}\\ &\displaystyle=\frac{1}{2}(2x-3)^{-7}\cdot 2&\mbox{}\\ &\displaystyle=(2x-3)^{-7}&\mbox{}\\ \end{array} $$
Dead Giveaway: If somewhere in the integrand you see a function and a multiple of its derivative, there's a good chance you have a substitution on your hands.
Integration by Substitution Strategy
1. Choose an expression $g(x)$ within the integrand (an "inside" function) to set equal to $u$ in such a way that that $g'(x)$ (the "outside" function) is also part of the integrand.
2. Substitute $u = g(x)$ and $du = g'(x)dx$ into the integral.
3. We should now be able to evaluate the integral with respect to $u.$ If the integral can’t be evaluated we need to go back and select a different expression to use as $u.$
4. Find the integral in terms of $u.$
5. Translate the result back to $x.$
Example
Use a suitable change of variables to determine the indefinite integral $$\int \cos^3 \theta \sin \theta \, d\theta$$ Note: The "outside" function $\sin \theta$ is a constant multiple of the derivative of the "inside" function $\cos \theta.$ This one of the dead giveaways that substitution will work.
Let $u=\cos \theta.$ Then $\displaystyle \frac{du}{d\theta}=-\sin \theta$ so that $\displaystyle -du=\sin \theta\,d\theta.$
Thus,
$$
\begin{array}{lll}
\displaystyle \int \cos^3 \theta \sin \theta \, d\theta&=\displaystyle \int u^3\cdot (-du)&\mbox{}\\
&=\displaystyle -\int u^3\,du&\mbox{}\\
&=\displaystyle -\frac{1}{4} u^4+C&\mbox{}\\
&=\displaystyle -\frac{1}{4} \cos^4 \theta+C&\mbox{}\\
\end{array}
$$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{d\theta}\left(-\frac{1}{4} \cos^4 \theta+C\right) &\displaystyle=-\frac{1}{4} \frac{d}{d\theta}\cos^4 \theta+\frac{d}{d\theta}C&\mbox{}\\ &\displaystyle=-\frac{1}{4}(4)\cos^3 \theta \frac{d}{d\theta} \cos \theta+0&\mbox{}\\ &\displaystyle=-\cos^3 \theta (-\sin \theta)&\mbox{}\\ &\displaystyle=\cos^3 \theta \sin \theta&\mbox{}\\ \end{array} $$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{d\theta}\left(-\frac{1}{4} \cos^4 \theta+C\right) &\displaystyle=-\frac{1}{4} \frac{d}{d\theta}\cos^4 \theta+\frac{d}{d\theta}C&\mbox{}\\ &\displaystyle=-\frac{1}{4}(4)\cos^3 \theta \frac{d}{d\theta} \cos \theta+0&\mbox{}\\ &\displaystyle=-\cos^3 \theta (-\sin \theta)&\mbox{}\\ &\displaystyle=\cos^3 \theta \sin \theta&\mbox{}\\ \end{array} $$
Example with a Definite Integral
Use a suitable change of variables to evaluate the definite integral $$\int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx$$ Note: The "outside" function $x$ is a constant multiple of the derivative of the "inside" function $1+x^2.$ This one of the dead giveaways that substitution will work.
We first find an antiderivative and then evaluate the definite integral.
Let $u=1+x^2.$ Then $\displaystyle \frac{du}{dx}=2x$ so that $\displaystyle \frac{1}{2}du=x\,dx.$ Thus, $$ \begin{array}{lll} \displaystyle \int \frac{x}{\sqrt{1+x^2}} \, dx&=\displaystyle \int \frac{x\,dx}{\sqrt{1+x^2}}&\mbox{}\\ &=\displaystyle \int \frac{\frac{1}{2}du}{\sqrt{u}}&\mbox{}\\ &=\displaystyle \frac{1}{2}\int u^{-1/2} du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left( 2u^{1/2}\right)+C&\mbox{}\\ &=\displaystyle \frac{1}{2}\left( 2\sqrt{u}\right)+C&\mbox{}\\ &=\displaystyle \sqrt{u}+C&\mbox{}\\ &=\displaystyle \sqrt{1+x^2}+C&\mbox{}\\ \end{array} $$ Then, $$\int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx=\left[\sqrt{1+x^2}\right]_{0}^{1}=\sqrt{1+1^2}-\sqrt{1+0^2}=\sqrt{2}-1$$ Special Note: Any antiderivative will work. In the above we chose $\sqrt{1+x^2},$ but notice that $\sqrt{1+x^2}+5$ would also work since we subtract.
Bonus: Scenic Alternative
We can reduce the number of steps by constructing a definite integral involving $u$ instead of $x.$
Using the same substitutions as above: $$ \begin{array}{lll} \displaystyle \int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx&=\displaystyle \int_{0}^{1} \frac{x\,dx}{\sqrt{1+x^2}}&\mbox{}\\ &=\displaystyle \int_{1}^{2} \frac{\frac{1}{2}du}{\sqrt{u}}&\mbox{when $x=0,$ $u=1$ and when $x=1,$ $u=2$}\\ &=\displaystyle \frac{1}{2}\int_{1}^{2} u^{-1/2} du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left[ 2u^{1/2}\right]_{1}^{2} du&\mbox{any antiderivative will work}\\ &=\displaystyle \frac{1}{2}\left[ 2\sqrt{u}\right]_{1}^{2} du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left[ 2\sqrt{2}-2\sqrt{1}\right]&\mbox{}\\ &=\displaystyle \sqrt{2}-\sqrt{1}&\mbox{}\\ &=\displaystyle \sqrt{2}-1&\mbox{}\\ \end{array} $$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\sqrt{1+x^2}+C\right) &\displaystyle=\frac{d}{dx}\sqrt{1+x^2}+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}(1+x^2)+0&\mbox{}\\ &\displaystyle=\frac{1}{2\sqrt{1+x^2}} (2x)&\mbox{}\\ &\displaystyle=\frac{x}{\sqrt{1+x^2}}&\mbox{}\\ \end{array} $$
Let $u=1+x^2.$ Then $\displaystyle \frac{du}{dx}=2x$ so that $\displaystyle \frac{1}{2}du=x\,dx.$ Thus, $$ \begin{array}{lll} \displaystyle \int \frac{x}{\sqrt{1+x^2}} \, dx&=\displaystyle \int \frac{x\,dx}{\sqrt{1+x^2}}&\mbox{}\\ &=\displaystyle \int \frac{\frac{1}{2}du}{\sqrt{u}}&\mbox{}\\ &=\displaystyle \frac{1}{2}\int u^{-1/2} du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left( 2u^{1/2}\right)+C&\mbox{}\\ &=\displaystyle \frac{1}{2}\left( 2\sqrt{u}\right)+C&\mbox{}\\ &=\displaystyle \sqrt{u}+C&\mbox{}\\ &=\displaystyle \sqrt{1+x^2}+C&\mbox{}\\ \end{array} $$ Then, $$\int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx=\left[\sqrt{1+x^2}\right]_{0}^{1}=\sqrt{1+1^2}-\sqrt{1+0^2}=\sqrt{2}-1$$ Special Note: Any antiderivative will work. In the above we chose $\sqrt{1+x^2},$ but notice that $\sqrt{1+x^2}+5$ would also work since we subtract.
Bonus: Scenic Alternative
We can reduce the number of steps by constructing a definite integral involving $u$ instead of $x.$
Using the same substitutions as above: $$ \begin{array}{lll} \displaystyle \int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx&=\displaystyle \int_{0}^{1} \frac{x\,dx}{\sqrt{1+x^2}}&\mbox{}\\ &=\displaystyle \int_{1}^{2} \frac{\frac{1}{2}du}{\sqrt{u}}&\mbox{when $x=0,$ $u=1$ and when $x=1,$ $u=2$}\\ &=\displaystyle \frac{1}{2}\int_{1}^{2} u^{-1/2} du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left[ 2u^{1/2}\right]_{1}^{2} du&\mbox{any antiderivative will work}\\ &=\displaystyle \frac{1}{2}\left[ 2\sqrt{u}\right]_{1}^{2} du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left[ 2\sqrt{2}-2\sqrt{1}\right]&\mbox{}\\ &=\displaystyle \sqrt{2}-\sqrt{1}&\mbox{}\\ &=\displaystyle \sqrt{2}-1&\mbox{}\\ \end{array} $$
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\sqrt{1+x^2}+C\right) &\displaystyle=\frac{d}{dx}\sqrt{1+x^2}+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}(1+x^2)+0&\mbox{}\\ &\displaystyle=\frac{1}{2\sqrt{1+x^2}} (2x)&\mbox{}\\ &\displaystyle=\frac{x}{\sqrt{1+x^2}}&\mbox{}\\ \end{array} $$
Words of Wisdom
After making your substitution, don't forget that an antiderivative must be in the original variable, not $u.$
On the other hand, if you're evaluating a definite integral and you change the limits of integration to fit $u$ like in the scenic alternative above, then there is no need to restate the antiderivative in terms of the original variable.
A Less-Obvious Substitution
Here, the "inside function" and its derivative hanging out somewhere isn't necessarily obvious. $$\int_{-1}^{1} z\sqrt{1-z} \, dz$$
Let $u=1-z.$ Then $du=-dz,$ or $\displaystyle \frac{1}{2}\,dz=-\,du.$
From the above we also have that $z=1-u.$
We will also transform the limits of integration: when $z=-1,$ $u=1-z=1-(-1)=2,$ and when $z=1,$ $u=1-z=1-1=0.$
We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int_{-1}^{1} z\sqrt{1-z} \, dz&=\displaystyle \int_{2}^{0} (1-u)\sqrt{u} (\, -du)&\mbox{}\\ &=\displaystyle -\int_{2}^{0} (1-u)\sqrt{u}\, du&\mbox{}\\ &=\displaystyle \int_{0}^{2} (1-u)\sqrt{u}\, du&\mbox{recall that $\displaystyle \int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx$}\\ &=\displaystyle \int_{0}^{2} u^{1/2}-u^{3/2}\, du&\mbox{}\\ &=\displaystyle \left[\frac{2}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right]_{0}^{2}&\mbox{}\\ &=\displaystyle \left[\frac{2}{3}(2)^{3/2}-\frac{2}{5}(2)^{5/2}\right]-\left[\frac{2}{3}(0)^{3/2}-\frac{2}{5}(0)^{5/2}\right]&\mbox{}\\ &=\displaystyle \left[\frac{2}{3}(\sqrt{2})^{3}-\frac{2}{5}(\sqrt{2})^{5}\right]-0&\mbox{}\\ &=\displaystyle \frac{2}{3}(2\sqrt{2})-\frac{2}{5}(4\sqrt{2})&\mbox{}\\ &=\displaystyle \frac{4}{3}\sqrt{2}-\frac{8}{5}\sqrt{2}&\mbox{}\\ &=\displaystyle \left(\frac{4}{3}-\frac{8}{5}\right)\sqrt{2}&\mbox{}\\ &=\displaystyle -\frac{4}{15}\sqrt{2}&\mbox{}\\ \end{array} $$
From the above we also have that $z=1-u.$
We will also transform the limits of integration: when $z=-1,$ $u=1-z=1-(-1)=2,$ and when $z=1,$ $u=1-z=1-1=0.$
We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int_{-1}^{1} z\sqrt{1-z} \, dz&=\displaystyle \int_{2}^{0} (1-u)\sqrt{u} (\, -du)&\mbox{}\\ &=\displaystyle -\int_{2}^{0} (1-u)\sqrt{u}\, du&\mbox{}\\ &=\displaystyle \int_{0}^{2} (1-u)\sqrt{u}\, du&\mbox{recall that $\displaystyle \int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx$}\\ &=\displaystyle \int_{0}^{2} u^{1/2}-u^{3/2}\, du&\mbox{}\\ &=\displaystyle \left[\frac{2}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right]_{0}^{2}&\mbox{}\\ &=\displaystyle \left[\frac{2}{3}(2)^{3/2}-\frac{2}{5}(2)^{5/2}\right]-\left[\frac{2}{3}(0)^{3/2}-\frac{2}{5}(0)^{5/2}\right]&\mbox{}\\ &=\displaystyle \left[\frac{2}{3}(\sqrt{2})^{3}-\frac{2}{5}(\sqrt{2})^{5}\right]-0&\mbox{}\\ &=\displaystyle \frac{2}{3}(2\sqrt{2})-\frac{2}{5}(4\sqrt{2})&\mbox{}\\ &=\displaystyle \frac{4}{3}\sqrt{2}-\frac{8}{5}\sqrt{2}&\mbox{}\\ &=\displaystyle \left(\frac{4}{3}-\frac{8}{5}\right)\sqrt{2}&\mbox{}\\ &=\displaystyle -\frac{4}{15}\sqrt{2}&\mbox{}\\ \end{array} $$
Sometimes More than One Substitution will Work
Use a suitable change of variables to determine the indefinite integral $$\int \frac{x^3}{\sqrt{x^2+1}} \, dx$$
Here we will find the integral twice using two separate substitutions.
Substitution 1 (More Obvious): Let $u=x^2+1.$ Then $du=2x\,dx,$ or $\displaystyle \frac{1}{2}\,du=x \,dx.$
From the above we also have that $x^2=u-1.$
We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx&=\displaystyle \int \frac{x^2}{\sqrt{x^2+1}} x\, dx&\mbox{}\\ &=\displaystyle \int \frac{u-1}{\sqrt{u}} \frac{1}{2}\,du&\mbox{}\\ &=\displaystyle \frac{1}{2}\int \frac{u-1}{\sqrt{u}} \,du&\mbox{}\\ &=\displaystyle \frac{1}{2}\int u^{1/2}-u^{-1/2} \,du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left(\frac{2}{3}u^{3/2}-2u^{1/2}\right)+C&\mbox{}\\ &=\displaystyle \frac{1}{3}u^{3/2}-u^{1/2}+C&\mbox{}\\ &=\displaystyle \frac{1}{3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C&\mbox{}\\ &=\displaystyle \frac{1}{3}\left(\sqrt{x^2+1}\right)^{3}-\sqrt{x^2+1}+C&\mbox{}\\ \end{array} $$
Substitution 2 (Less Obvious): Let $u=\sqrt{x^2+1}.$ Then $\displaystyle du=\frac{x\,dx}{x^2+1}.$ From the above we also have that $x^2=u^2-1.$ We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx&=\displaystyle \int x^2 \frac{x}{\sqrt{x^2+1}} \, dx&\mbox{}\\ &=\displaystyle \int (u^2-1) du&\mbox{}\\ &=\displaystyle \frac{1}{3}u^3-u+C&\mbox{}\\ &=\displaystyle \frac{1}{3}\left(\sqrt{x^2+1}\right)^3-\sqrt{x^2+1}+C&\mbox{}\\ \end{array} $$
Clearly, some substitutions are less work than others!
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\frac{1}{3}(\sqrt{x^2+1})^3-\sqrt{x^2+1}+C\right) &\displaystyle=\frac{1}{3}\frac{d}{dx}\left(\sqrt{x^2+1}\right)^3-\frac{d}{dx}\sqrt{x^2+1}+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{3}\cdot 3\left(\sqrt{x^2+1}\right)^2\frac{d}{dx}\sqrt{x^2+1}-\frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}(x^2+1)+0&\mbox{}\\ &\displaystyle=\left(\sqrt{x^2+1}\right)^2\frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}(x^2+1)-\frac{1}{2\sqrt{x^2+1}}\cdot 2x&\mbox{}\\ &\displaystyle=\frac{x^2+1}{2\sqrt{x^2+1}}\cdot 2x-\frac{x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{(x^2+1)x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{x^3+x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{x^3+x-x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{x^3}{\sqrt{x^2+1}}&\mbox{}\\ \end{array} $$
Substitution 1 (More Obvious): Let $u=x^2+1.$ Then $du=2x\,dx,$ or $\displaystyle \frac{1}{2}\,du=x \,dx.$
From the above we also have that $x^2=u-1.$
We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx&=\displaystyle \int \frac{x^2}{\sqrt{x^2+1}} x\, dx&\mbox{}\\ &=\displaystyle \int \frac{u-1}{\sqrt{u}} \frac{1}{2}\,du&\mbox{}\\ &=\displaystyle \frac{1}{2}\int \frac{u-1}{\sqrt{u}} \,du&\mbox{}\\ &=\displaystyle \frac{1}{2}\int u^{1/2}-u^{-1/2} \,du&\mbox{}\\ &=\displaystyle \frac{1}{2}\left(\frac{2}{3}u^{3/2}-2u^{1/2}\right)+C&\mbox{}\\ &=\displaystyle \frac{1}{3}u^{3/2}-u^{1/2}+C&\mbox{}\\ &=\displaystyle \frac{1}{3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C&\mbox{}\\ &=\displaystyle \frac{1}{3}\left(\sqrt{x^2+1}\right)^{3}-\sqrt{x^2+1}+C&\mbox{}\\ \end{array} $$
Substitution 2 (Less Obvious): Let $u=\sqrt{x^2+1}.$ Then $\displaystyle du=\frac{x\,dx}{x^2+1}.$ From the above we also have that $x^2=u^2-1.$ We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx&=\displaystyle \int x^2 \frac{x}{\sqrt{x^2+1}} \, dx&\mbox{}\\ &=\displaystyle \int (u^2-1) du&\mbox{}\\ &=\displaystyle \frac{1}{3}u^3-u+C&\mbox{}\\ &=\displaystyle \frac{1}{3}\left(\sqrt{x^2+1}\right)^3-\sqrt{x^2+1}+C&\mbox{}\\ \end{array} $$
Clearly, some substitutions are less work than others!
Differentiation Check: $$ \begin{array}{lll} \displaystyle \frac{d}{dx}\left(\frac{1}{3}(\sqrt{x^2+1})^3-\sqrt{x^2+1}+C\right) &\displaystyle=\frac{1}{3}\frac{d}{dx}\left(\sqrt{x^2+1}\right)^3-\frac{d}{dx}\sqrt{x^2+1}+\frac{d}{dx}C&\mbox{}\\ &\displaystyle=\frac{1}{3}\cdot 3\left(\sqrt{x^2+1}\right)^2\frac{d}{dx}\sqrt{x^2+1}-\frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}(x^2+1)+0&\mbox{}\\ &\displaystyle=\left(\sqrt{x^2+1}\right)^2\frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}(x^2+1)-\frac{1}{2\sqrt{x^2+1}}\cdot 2x&\mbox{}\\ &\displaystyle=\frac{x^2+1}{2\sqrt{x^2+1}}\cdot 2x-\frac{x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{(x^2+1)x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{x^3+x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{x^3+x-x}{\sqrt{x^2+1}}&\mbox{}\\ &\displaystyle=\frac{x^3}{\sqrt{x^2+1}}&\mbox{}\\ \end{array} $$
Bonus Example
The area of a semicircle of radius $1$ can be expressed as $\displaystyle \int_{-1}^{1} \sqrt{1-x^2} \, dx.$ Use the substitution $x=\cos t$ to express the area of the semicircle as an integral of a trigonmetric function. You do not need to compute the integral.
Let $x=\cos t$ so that $dx=-\sin t \,dt$
We will also transform the limits of integration. When $x=-1,$ $-1=\cos t$ which gives that $t=\pi.$ Similarly, when $x=1,$ $1=\cos t$ which gives that $t=0.$
We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int_{-1}^{1} \sqrt{1-x^2} \, dx&=\displaystyle \int_{\pi}^{0} \sqrt{1-\cos^2 t} (-\sin t \,dt)&\mbox{}\\ &=\displaystyle -\int_{\pi}^{0} \sqrt{1-\cos^2 t} \sin t \,dt&\mbox{}\\ &=\displaystyle \int_{0}^{\pi} \sqrt{1-\cos^2 t} \sin t \,dt&\mbox{recall that $\displaystyle \int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx$}\\ &=\displaystyle \int_{0}^{\pi} \sqrt{\sin^2 t} \sin t \,dt&\mbox{}\\ &=\displaystyle \int_{0}^{\pi} |\sin t| \sin t \,dt&\mbox{}\\ &=\displaystyle \int_{0}^{\pi} \sin t \cdot \sin t \,dt&\mbox{$|\sin t|=\sin t$ since $\sin t \geq 0$ on $0 \leq t \leq \pi$}\\ &=\displaystyle \int_{0}^{\pi} \sin^2 t \,dt&\mbox{}\\ \end{array} $$
We will also transform the limits of integration. When $x=-1,$ $-1=\cos t$ which gives that $t=\pi.$ Similarly, when $x=1,$ $1=\cos t$ which gives that $t=0.$
We may now transform the original integral: $$ \begin{array}{lll} \displaystyle \int_{-1}^{1} \sqrt{1-x^2} \, dx&=\displaystyle \int_{\pi}^{0} \sqrt{1-\cos^2 t} (-\sin t \,dt)&\mbox{}\\ &=\displaystyle -\int_{\pi}^{0} \sqrt{1-\cos^2 t} \sin t \,dt&\mbox{}\\ &=\displaystyle \int_{0}^{\pi} \sqrt{1-\cos^2 t} \sin t \,dt&\mbox{recall that $\displaystyle \int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx$}\\ &=\displaystyle \int_{0}^{\pi} \sqrt{\sin^2 t} \sin t \,dt&\mbox{}\\ &=\displaystyle \int_{0}^{\pi} |\sin t| \sin t \,dt&\mbox{}\\ &=\displaystyle \int_{0}^{\pi} \sin t \cdot \sin t \,dt&\mbox{$|\sin t|=\sin t$ since $\sin t \geq 0$ on $0 \leq t \leq \pi$}\\ &=\displaystyle \int_{0}^{\pi} \sin^2 t \,dt&\mbox{}\\ \end{array} $$