U-Substitution Integration

Introduction to U-Substitution	Which Method to Use?
U-Substitution Integration Problems	More Practice

Note that U-Substitution with Definite Integration can be found here in the Definite Integration section, U-Substitution with Exponential and Logarithmic Integration can be found in the Exponential and Logarithmic Integration section, and U-Substitution with Inverse Trig Functions can be found in the Derivatives and Integrals of Inverse Trig Functions section.

Introduction to U-Substitution

U-Substitution Integration, or U-Sub Integration, is the opposite of the The Chain Rule from Differential Calculus, but it’s a little trickier since you have to set it up like a puzzle. Once you get the hang of it, it’s fun, though! U-sub is also known the reverse chain rule or change of variables.

Why do we have to do something other than just integrate like we learned? Basically, we need U-sub to take the anti-derivative of a Composite Function; it’s the “undoing” of the Chain Rule. For example, for $ \displaystyle \frac{{d\left( {\sin \left( {4x} \right)} \right)}}{{dx}}=4\cos \left( {4x} \right)$, we multiplied our “$ \cos \left( {4x} \right)$” answer by 4, since the derivative of $ 4x$ is 4. When we go backward and integrate, it turns out we will have to divide by 4. But it’s not always that easy, so we’ll learn some techniques to do the substitution.

The reason the technique is called “U-substitution” is because we substitute the more complicated expression (like “$ 4x$” above) with a $ u$ (a simple variable), do the integration, and then substitute back the more complicated expression. The “$ u$” can be thought of as the “inside” function. This is also called “change of variables”.

And remember to always take the derivative back (if you have time!) to make sure you’ve done the problem correctly. Differentiation tends to be a little easier than integration. Here is a more “formal” definition; don’t worry if you don’t get this; you won’t 🙂 :

U-Substitution Integration

Outside Function Derivative of Inside Function

$ \displaystyle \int{{\color{red}{{f\left( {\color{green}{{g\left( x \right)}}} \right)}}\,\color{purple}{{{g}’\left( x \right)}}}}\,dx=\color{blue}{{F\left( {g\left( x \right)} \right)}}\,+\,C$

Inside Function Any Composite Function

What this says is that if we want the integral of the outside function, to make it work, we have to make sure that what we’re integrating somehow has a factor that is the derivative of the inside function. (We can “trick” the integrand into having this factor.)

U-Substitution Integration Problems

Let’s do some problems and set up the $ u$-sub. The trickiest thing is probably to know what to use as the $ u$ (the inside function); this is typically an expression that you are raising to a power, taking a trig function of, and so on, when it’s not just an “$ x$”. Sometimes, but not always, it’s the function inside parentheses or under a radical sign.

In the following problems, we are “lucky”; the problems fit into the mold of a “chain rule” problem, so we can easily do the integration. Again, it’s a good idea to take the derivative back from the answer to make sure we get the integral!

$ \boldsymbol {U}$-Sub Integral: $ \displaystyle \int{{f\left( {g\left( x \right)} \right)\,}}{g}’\left( x \right)dx$	$ u=g\left( x \right)$	$ \displaystyle du={g}’\left( x \right)dx$	Substitute, Integrate, and Substitute Back!
$ \displaystyle \int{{{{{\left( {\color{blue}{{5{{x}^{2}}+3}}} \right)}}^{4}}}}\color{red}{{\left( {10x} \right)dx}}$	$ u=5{{x}^{2}}+3$	$ \displaystyle du=\left( {10x} \right)dx$	$ \displaystyle \int{{{{{\color{blue}{{\left( {5{{x}^{2}}+3} \right)}}}}^{4}}}}\color{red}{{\left( {10x} \right)dx}}=\int{{{{{\color{blue}{u}}}^{4}}}}\,\color{red}{{du}}=\frac{1}{5}{{u}^{5}}+C=\frac{1}{5}{{\left( {5{{x}^{2}}+3} \right)}^{5}}+C$
$ \displaystyle \int{{\color{red}{{{{{\sec }}^{2}}x}}\sqrt{{\color{blue}{{\tan x}}}}}}\,\color{red}{{dx}}$	$ u=\tan x$	$ du=\left( {{{{\sec }}^{2}}x\,} \right)dx$	$ \displaystyle \int{{\color{red}{{{{{\sec }}^{2}}x}}\sqrt{{\color{blue}{{\tan x}}}}}}\,\color{red}{{dx}}=\int{{\sqrt{{\color{blue}{u}}}\,}}\color{red}{{du}}=\int{{{{{\color{blue}{u}}}^{{\frac{1}{2}}}}}}\,\color{red}{{du}}=\frac{2}{3}{{u}^{{\frac{3}{2}}}}+C=\frac{2}{3}{{\left( {\tan x} \right)}^{{\frac{3}{2}}}}+C$

Most $ u$-sub problems won’t work exactly like this though; with most $ u$-sub problems, we have to somehow get rid of the “extra” variables in the problem by solving for $ dx$ and canceling them out. If we can do this (sometimes we can’t!), we can solve with $ u$-sub. I like to organize the substitutions like this, to really show what’s going on.

Note how we start by picking the “$ u$” to be the expression that is raised to a power, or that we take a root of, or is the argument of a trig function (for example, “$ 4x$”):

$ \boldsymbol {U}$-Sub Integral	Setup	Solution and Check
$ \displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx$	$ \displaystyle u=2{{x}^{2}}+3$ $ \displaystyle du=4x\,dx$ $ \displaystyle dx=\frac{{du}}{{4x}}$	$ \require {cancel} \displaystyle \int{{{{{\left( {\color{blue}{{2{{x}^{2}}+3}}} \right)}}^{5}}}}x\,\color{green}{{dx}}=\int{{{{{\color{blue}{u}}}^{5}}}}\cancel{x}\color{green}{{\frac{{du}}{{4\cancel{x}}}}}=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{5}}\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{6}{{\color{blue}{u}}^{6}}+C=\frac{1}{{24}}{{\color{blue}{u}}^{6}}+C$ $ \displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx=\frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{6}}+C=\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}+C$ Differentiate back to verify: $ \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}} \right)&=6\cdot \frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot 4x\\&=\frac{1}{{\cancel{4}}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot \cancel{4}x={{\left( {2{{x}^{2}}+3} \right)}^{5}}x\end{align}$ √
$ \int{{\cos \left( {4x} \right)\,{{{\sin }}^{3}}}}\left( {4x} \right)\,dx$ Note that this is actually: $ \displaystyle \int{{\cos \left( {4x} \right)\,}}{{\left( {\sin \left( {4x} \right)} \right)}^{3}}dx$	$ \displaystyle u=\sin \left( {4x} \right)$ $ \displaystyle du=4\cos \left( {4x} \right)\,dx$ $ \displaystyle dx=\frac{{du}}{{4\cos \left( {4x} \right)}}$	$ \displaystyle \begin{align}\int{{\cos \left( {4x} \right){{{\color{blue}{{\sin }}}}^{3}}\left( {\color{blue}{{4x}}} \right)}}\,\color{green}{{dx}}&=\int{{\cancel{{\cos \left( {4x} \right)}}{{{\color{blue}{u}}}^{3}}}}\,\color{green}{{\frac{{du}}{{4\cancel{{\cos \left( {4x} \right)}}}}}}=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{3}}\,\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{4}{{\color{blue}{u}}^{4}}+C=\frac{1}{{16}}{{\color{blue}{u}}^{4}}+C\end{align}$ $ \displaystyle \int{{\cos \left( {4x} \right){{{\sin }}^{3}}\left( {4x} \right)}}\,dx=\frac{1}{{16}}{{\sin }^{4}}\left( {4x} \right)+C$ Differentiate back to verify: $ \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{1}{{16}}{{{\sin }}^{4}}\left( {4x} \right)} \right)&=\frac{1}{{16}}\cdot 4\sin {{\left( {4x} \right)}^{3}}\cdot \cos \left( {4x} \right)\cdot 4\\&=\cancel{{\frac{{16}}{{16}}}}\sin {{\left( {4x} \right)}^{3}}\cos \left( {4x} \right)=\cos \left( {4x} \right){{\sin }^{3}}\left( {4x} \right)\end{align}$ √
$ \displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}$	$ \displaystyle u=3{{x}^{2}}+2$ $ \displaystyle du=6x\,dx$ $ \displaystyle dx=\frac{{du}}{{6x}}$	$ \displaystyle \begin{align} \int{{\frac{{x\,\color{green}{dx}}}{{\sqrt{{\color{blue}{{3{{x}^{2}}+2}}}}}}}}&=\int{{{{{\left( {\color{blue}{{3{{x}^{2}}+2}}} \right)}}^{{-\frac{1}{2}}}}x\,\color{green}{{dx}}=}}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}}}\cancel{x}\color{green}{{\frac{{du}}{{6\cancel{x}}}}}=\frac{1}{6}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}\color{green}{du}}}=\frac{{\left( {\frac{1}{6}} \right)}}{{\left( {\frac{1}{2}} \right)}}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C=\frac{1}{3}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C\end{align}$ $ \displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}x\,dx=\frac{1}{3}{{\left( {3{{x}^{2}}+2} \right)}^{{\frac{1}{2}}}}+C=\frac{1}{3}\sqrt{{3{{x}^{2}}+2}}\,+\,C$ Differentiate back to verify: $ \displaystyle \frac{d}{{dx}}\left( {\frac{1}{3}\sqrt{{3{{x}^{2}}+2}}} \right)\,=\frac{d}{{dx}}\left( {\frac{1}{3}{{{\left( {3{{x}^{2}}+2} \right)}}^{{\frac{1}{2}}}}} \right)=\frac{1}{{\cancel{6}}}{{\left( {3{{x}^{2}}+2} \right)}^{{-\frac{1}{2}}}}\cdot \cancel{6}x=\frac{x}{{\sqrt{{3{{x}^{2}}+2}}}}$ √

$ \boldsymbol {U}$-Sub Integral

Setup

Solution and Check

$ \displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx$

$ \displaystyle u=2{{x}^{2}}+3$

$ \displaystyle du=4x\,dx$

$ \displaystyle dx=\frac{{du}}{{4x}}$

$ \require {cancel} \displaystyle \int{{{{{\left( {\color{blue}{{2{{x}^{2}}+3}}} \right)}}^{5}}}}x\,\color{green}{{dx}}=\int{{{{{\color{blue}{u}}}^{5}}}}\cancel{x}\color{green}{{\frac{{du}}{{4\cancel{x}}}}}=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{5}}\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{6}{{\color{blue}{u}}^{6}}+C=\frac{1}{{24}}{{\color{blue}{u}}^{6}}+C$

$ \displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx=\frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{6}}+C=\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}+C$

Differentiate back to verify:

$ \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}} \right)&=6\cdot \frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot 4x\\&=\frac{1}{{\cancel{4}}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot \cancel{4}x={{\left( {2{{x}^{2}}+3} \right)}^{5}}x\end{align}$ √

$ \int{{\cos \left( {4x} \right)\,{{{\sin }}^{3}}}}\left( {4x} \right)\,dx$

Note that this is actually:

$ \displaystyle \int{{\cos \left( {4x} \right)\,}}{{\left( {\sin \left( {4x} \right)} \right)}^{3}}dx$

$ \displaystyle u=\sin \left( {4x} \right)$

$ \displaystyle du=4\cos \left( {4x} \right)\,dx$

$ \displaystyle dx=\frac{{du}}{{4\cos \left( {4x} \right)}}$

$ \displaystyle \begin{align}\int{{\cos \left( {4x} \right){{{\color{blue}{{\sin }}}}^{3}}\left( {\color{blue}{{4x}}} \right)}}\,\color{green}{{dx}}&=\int{{\cancel{{\cos \left( {4x} \right)}}{{{\color{blue}{u}}}^{3}}}}\,\color{green}{{\frac{{du}}{{4\cancel{{\cos \left( {4x} \right)}}}}}}=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{3}}\,\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{4}{{\color{blue}{u}}^{4}}+C=\frac{1}{{16}}{{\color{blue}{u}}^{4}}+C\end{align}$

$ \displaystyle \int{{\cos \left( {4x} \right){{{\sin }}^{3}}\left( {4x} \right)}}\,dx=\frac{1}{{16}}{{\sin }^{4}}\left( {4x} \right)+C$

Differentiate back to verify:

$ \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{1}{{16}}{{{\sin }}^{4}}\left( {4x} \right)} \right)&=\frac{1}{{16}}\cdot 4\sin {{\left( {4x} \right)}^{3}}\cdot \cos \left( {4x} \right)\cdot 4\\&=\cancel{{\frac{{16}}{{16}}}}\sin {{\left( {4x} \right)}^{3}}\cos \left( {4x} \right)=\cos \left( {4x} \right){{\sin }^{3}}\left( {4x} \right)\end{align}$ √

$ \displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}$

$ \displaystyle u=3{{x}^{2}}+2$

$ \displaystyle du=6x\,dx$

$ \displaystyle dx=\frac{{du}}{{6x}}$

$ \displaystyle \begin{align} \int{{\frac{{x\,\color{green}{dx}}}{{\sqrt{{\color{blue}{{3{{x}^{2}}+2}}}}}}}}&=\int{{{{{\left( {\color{blue}{{3{{x}^{2}}+2}}} \right)}}^{{-\frac{1}{2}}}}x\,\color{green}{{dx}}=}}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}}}\cancel{x}\color{green}{{\frac{{du}}{{6\cancel{x}}}}}=\frac{1}{6}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}\color{green}{du}}}=\frac{{\left( {\frac{1}{6}} \right)}}{{\left( {\frac{1}{2}} \right)}}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C=\frac{1}{3}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C\end{align}$

$ \displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}x\,dx=\frac{1}{3}{{\left( {3{{x}^{2}}+2} \right)}^{{\frac{1}{2}}}}+C=\frac{1}{3}\sqrt{{3{{x}^{2}}+2}}\,+\,C$

Differentiate back to verify:

$ \displaystyle \frac{d}{{dx}}\left( {\frac{1}{3}\sqrt{{3{{x}^{2}}+2}}} \right)\,=\frac{d}{{dx}}\left( {\frac{1}{3}{{{\left( {3{{x}^{2}}+2} \right)}}^{{\frac{1}{2}}}}} \right)=\frac{1}{{\cancel{6}}}{{\left( {3{{x}^{2}}+2} \right)}^{{-\frac{1}{2}}}}\cdot \cancel{6}x=\frac{x}{{\sqrt{{3{{x}^{2}}+2}}}}$ √

One thing that we can notice from the above is a $ u$-sub simplification formula we can use, although it’s still good to know the mechanics on how to do the $ u$-sub integration. But here goes:

U-Substitution Integration Simplification Formula

$ \displaystyle \int{{{{{\left( {ax+b} \right)}}^{n}}\,dx=\frac{{{{{\left( {ax+b} \right)}}^{{n+1}}}}}{{a\left( {n+1} \right)}}}}+\,C$

Here are a few more $ u$-sub trig problems. Notice in the first problem, we have to separate the $ {{\sec }^{4}}x\cdot \tan x$ to $ {{\sec }^{3}}x\cdot \sec x\cdot \tan x$ so we can perform the $ u$-sub.

$ \boldsymbol {U}$-Sub Integral	Setup	Solution and Check
$ \displaystyle \int{{{{{\sec }}^{4}}\left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)}}\,dx$ Note that this is actually: $ \displaystyle \int{{{{{\left( {\sec \left( {\frac{x}{4}} \right)} \right)}}^{4}}\tan \left( {\frac{x}{4}} \right)}}\,dx$	$ \displaystyle u=\sec \left( {\frac{x}{4}} \right)$ $ \displaystyle \begin{align}du&=\sec \left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)\cdot \frac{1}{4}dx\\&=\frac{{\sec \left( {\frac{x}{4}} \right)\,\tan \left( {\frac{x}{4}} \right)}}{4}\,dx\end{align}$ $ \displaystyle dx=\frac{4}{{\sec \left( {\frac{x}{4}} \right)\,\tan \left( {\frac{x}{4}} \right)}}\,du$	$ \displaystyle \require {cancel} \begin{align}\int{{{{{\color{blue}{{\sec }}}}^{4}}\left( {\color{blue}{{\frac{x}{4}}}} \right)\tan \left( {\frac{x}{4}} \right)}}\,\color{green}{{dx}}&=\int{{{{{\color{blue}{{\sec }}}}^{3}}\left( {\color{blue}{{\frac{x}{4}}}} \right)\cancel{{\sec \left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)}}}}\,\cdot \color{green}{{\frac{4}{{\cancel{{\sec \left( {\frac{x}{4}} \right)\,\tan \left( {\frac{x}{4}} \right)}}}}\,du}}\\&=4\int{{{{{\color{blue}{u}}}^{3}}}}\,\color{green}{{du}}=\color{green}{4}\cdot \frac{{{{{\color{blue}{u}}}^{4}}}}{{\cancel{4}}}+C={{\color{blue}{u}}^{4}}+C\end{align}$ $ \displaystyle \int{{{{{\sec }}^{4}}\left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)}}\,dx={{\sec }^{4}}\left( {\frac{x}{4}} \right)+C$ Differentiate back to verify: $ \displaystyle \begin{align}\frac{d}{{dx}}\left( {{{{\sec }}^{4}}\left( {\frac{x}{4}} \right)} \right)&=\cancel{4}{{\sec }^{3}}\left( {\frac{x}{4}} \right)\cdot \sec \left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)\frac{1}{{\cancel{4}}}\\&={{\sec }^{4}}\left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)\,\end{align}$ √
$ \displaystyle \int{{\frac{{\sin \left( {\sqrt{x}} \right)}}{{\sqrt{x}}}}}\,dx$	$ \displaystyle u=\sqrt{x}={{x}^{{\frac{1}{2}}}}$ $ \displaystyle du=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}dx$ $ \displaystyle dx=\frac{{du}}{{\frac{1}{2}{{x}^{{-\frac{1}{2}}}}}}=2{{x}^{{\frac{1}{2}}}}\,du$	$ \displaystyle \begin{align}\int{{\frac{{\sin \left( {\color{blue}{{\sqrt{x}}}} \right)}}{{\sqrt{x}}}}}\,\color{green}{{dx}}&=\int{{\sin \left( {\color{blue}{{\sqrt{x}}}} \right){{x}^{{-\frac{1}{2}}}}\,}}\color{green}{{dx}}=\int{{\sin \left( {\color{blue}{u}} \right){{x}^{{-\frac{1}{2}}}}\,}}\color{green}{{2{{x}^{{\frac{1}{2}}}}\,du}}=2\int{{\sin \left( {\color{blue}{u}} \right){{x}^{0}}\,}}\color{green}{{du}}\\&=2\int{{\sin \left( {\color{blue}{u}} \right)\,\color{green}{{du}}}}=-2\cos \left( u \right)\,+\,C=-2\cos \left( {\sqrt{x}} \right)\,+\,C\end{align}$ $ \displaystyle \int{{\frac{{\sin \left( {\sqrt{x}} \right)}}{{\sqrt{x}}}}}\,dx=-2\cos \left( {\sqrt{x}} \right)\,+\,C$ Differentiate back to verify: $ \displaystyle \begin{align}\frac{d}{{dx}}\left( {-2\cos \left( {\sqrt{x}} \right)} \right)&=\frac{d}{{dx}}\left( {-2\cos \left( {{{x}^{{\frac{1}{2}}}}} \right)} \right)\\&=-\cancel{2}\cdot -\sin \left( {{{x}^{{\frac{1}{2}}}}} \right)\cdot \cancel{{\frac{1}{2}}}{{x}^{{-\frac{1}{2}}}}=\frac{{\sin \left( {\sqrt{x}} \right)}}{{\sqrt{x}}}\end{align}$ √

$ \boldsymbol {U}$-Sub Integral

Setup

Solution and Check

$ \displaystyle \int{{{{{\sec }}^{4}}\left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)}}\,dx$

Note that this is actually:

$ \displaystyle \int{{{{{\left( {\sec \left( {\frac{x}{4}} \right)} \right)}}^{4}}\tan \left( {\frac{x}{4}} \right)}}\,dx$

$ \displaystyle u=\sec \left( {\frac{x}{4}} \right)$

$ \displaystyle \begin{align}du&=\sec \left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)\cdot \frac{1}{4}dx\\&=\frac{{\sec \left( {\frac{x}{4}} \right)\,\tan \left( {\frac{x}{4}} \right)}}{4}\,dx\end{align}$

$ \displaystyle dx=\frac{4}{{\sec \left( {\frac{x}{4}} \right)\,\tan \left( {\frac{x}{4}} \right)}}\,du$

$ \displaystyle \require {cancel} \begin{align}\int{{{{{\color{blue}{{\sec }}}}^{4}}\left( {\color{blue}{{\frac{x}{4}}}} \right)\tan \left( {\frac{x}{4}} \right)}}\,\color{green}{{dx}}&=\int{{{{{\color{blue}{{\sec }}}}^{3}}\left( {\color{blue}{{\frac{x}{4}}}} \right)\cancel{{\sec \left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)}}}}\,\cdot \color{green}{{\frac{4}{{\cancel{{\sec \left( {\frac{x}{4}} \right)\,\tan \left( {\frac{x}{4}} \right)}}}}\,du}}\\&=4\int{{{{{\color{blue}{u}}}^{3}}}}\,\color{green}{{du}}=\color{green}{4}\cdot \frac{{{{{\color{blue}{u}}}^{4}}}}{{\cancel{4}}}+C={{\color{blue}{u}}^{4}}+C\end{align}$

$ \displaystyle \int{{{{{\sec }}^{4}}\left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)}}\,dx={{\sec }^{4}}\left( {\frac{x}{4}} \right)+C$

Differentiate back to verify:

$ \displaystyle \begin{align}\frac{d}{{dx}}\left( {{{{\sec }}^{4}}\left( {\frac{x}{4}} \right)} \right)&=\cancel{4}{{\sec }^{3}}\left( {\frac{x}{4}} \right)\cdot \sec \left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)\frac{1}{{\cancel{4}}}\\&={{\sec }^{4}}\left( {\frac{x}{4}} \right)\tan \left( {\frac{x}{4}} \right)\,\end{align}$ √

$ \displaystyle \int{{\frac{{\sin \left( {\sqrt{x}} \right)}}{{\sqrt{x}}}}}\,dx$

$ \displaystyle u=\sqrt{x}={{x}^{{\frac{1}{2}}}}$

$ \displaystyle du=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}dx$

$ \displaystyle dx=\frac{{du}}{{\frac{1}{2}{{x}^{{-\frac{1}{2}}}}}}=2{{x}^{{\frac{1}{2}}}}\,du$

$ \displaystyle \begin{align}\int{{\frac{{\sin \left( {\color{blue}{{\sqrt{x}}}} \right)}}{{\sqrt{x}}}}}\,\color{green}{{dx}}&=\int{{\sin \left( {\color{blue}{{\sqrt{x}}}} \right){{x}^{{-\frac{1}{2}}}}\,}}\color{green}{{dx}}=\int{{\sin \left( {\color{blue}{u}} \right){{x}^{{-\frac{1}{2}}}}\,}}\color{green}{{2{{x}^{{\frac{1}{2}}}}\,du}}=2\int{{\sin \left( {\color{blue}{u}} \right){{x}^{0}}\,}}\color{green}{{du}}\\&=2\int{{\sin \left( {\color{blue}{u}} \right)\,\color{green}{{du}}}}=-2\cos \left( u \right)\,+\,C=-2\cos \left( {\sqrt{x}} \right)\,+\,C\end{align}$

$ \displaystyle \int{{\frac{{\sin \left( {\sqrt{x}} \right)}}{{\sqrt{x}}}}}\,dx=-2\cos \left( {\sqrt{x}} \right)\,+\,C$

Differentiate back to verify:

$ \displaystyle \begin{align}\frac{d}{{dx}}\left( {-2\cos \left( {\sqrt{x}} \right)} \right)&=\frac{d}{{dx}}\left( {-2\cos \left( {{{x}^{{\frac{1}{2}}}}} \right)} \right)\\&=-\cancel{2}\cdot -\sin \left( {{{x}^{{\frac{1}{2}}}}} \right)\cdot \cancel{{\frac{1}{2}}}{{x}^{{-\frac{1}{2}}}}=\frac{{\sin \left( {\sqrt{x}} \right)}}{{\sqrt{x}}}\end{align}$ √

Here’s one more problem when we have to solve for both $ dx$ in terms of $ du$ and $ x$ in terms of $ u$ to make the $ u$-sub work. Try doing something like this when you otherwise get stuck:

$ \boldsymbol {U}$-Sub Integral	Setup	Solution and Check
$ \int{{x\sqrt{{\,3x-1}}}}\,dx$	$ \displaystyle u=3x-1$ $ \displaystyle x=\frac{{u+1}}{3}$ $ \displaystyle du=3\,dx$ $ \displaystyle dx=\frac{{du}}{3}$	$ \require {cancel} \displaystyle \begin{align}\int{{\color{brown}{x}\sqrt{{\color{blue}{{3x-1}}}}}}\,\color{green}{{dx}}&=\int{{\color{brown}{x}{{{\left( {\color{blue}{{3x-1}}} \right)}}^{{\frac{1}{2}}}}\color{green}{{dx}}=\int{{\left( {\color{brown}{{\frac{{u+1}}{3}}}} \right){{{\color{blue}{u}}}^{{\frac{1}{2}}}}\color{green}{{dx}}=}}}}\int{{\left( {\color{brown}{{\frac{{u+1}}{3}}}} \right){{{\color{blue}{u}}}^{{\frac{1}{2}}}}\color{green}{{\frac{{du}}{3}}}}}\\&=\frac{1}{9}\int{{\left( {{{{\color{blue}{u}}}^{{\frac{3}{2}}}}+{{{\color{blue}{u}}}^{{\frac{1}{2}}}}} \right)\color{green}{{du}}}}=\frac{1}{9}\left( {\frac{{{{{\color{blue}{u}}}^{{\frac{5}{2}}}}}}{{\frac{5}{2}}}+\frac{{{{{\color{blue}{u}}}^{{\frac{3}{2}}}}}}{{\frac{3}{2}}}} \right)+C=\frac{2}{{45}}{{\color{blue}{u}}^{{\frac{5}{2}}}}+\frac{2}{{27}}{{\color{blue}{u}}^{{\frac{3}{2}}}}+C\\&=\frac{2}{{45}}{{\left( {3x-1} \right)}^{{\frac{5}{2}}}}+\frac{2}{{27}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}+C\end{align}$ If take out the Greatest Common Factor (GCF): $ \displaystyle =\frac{2}{{135}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}\left[ {3\left( {3x-1} \right)+5} \right]+C=\frac{2}{{135}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}\left( {9x+2} \right)+C=\frac{{2\sqrt{{{{{\left( {3x-1} \right)}}^{3}}}}\left( {9x+2} \right)}}{{135}}+C$ (Notice that $ 135$ is the Least Common Multiple (LCM) of $ 45$ and $ 27$) Differentiate back to verify: $ \require {cancel} \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{2}{{45}}{{{\left( {3x-1} \right)}}^{{\frac{5}{2}}}}+\frac{2}{{27}}{{{\left( {3x-1} \right)}}^{{\frac{3}{2}}}}} \right)&=\frac{{{}^{1}\cancel{5}}}{{\cancel{2}}}\cdot \frac{{\cancel{2}}}{{{{{\cancel{{45}}}}_{9}}}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}\cdot 3+\frac{{\cancel{2}}}{{{}_{9}\cancel{{27}}}}\cdot \frac{{{{{\cancel{3}}}^{1}}}}{{\cancel{2}}}{{\left( {3x-1} \right)}^{{\frac{1}{2}}}}\cdot 3\\&=\frac{1}{3}\left( {{{{\left( {3x-1} \right)}}^{{\frac{3}{2}}}}+{{{\left( {3x-1} \right)}}^{{\frac{1}{2}}}}} \right)=\frac{1}{3}{{\left( {3x-1} \right)}^{{\frac{1}{2}}}}\left( {3x-1+1} \right)\\&={{\left( {3x-1} \right)}^{{\frac{1}{2}}}}\cdot \frac{1}{{\cancel{3}}}\left( {\cancel{3}x} \right)=x\sqrt{{3x-1}}\end{align}$ √

$ \boldsymbol {U}$-Sub Integral

Setup

Solution and Check

$ \int{{x\sqrt{{\,3x-1}}}}\,dx$

$ \displaystyle u=3x-1$

$ \displaystyle x=\frac{{u+1}}{3}$

$ \displaystyle du=3\,dx$

$ \displaystyle dx=\frac{{du}}{3}$

$ \require {cancel} \displaystyle \begin{align}\int{{\color{brown}{x}\sqrt{{\color{blue}{{3x-1}}}}}}\,\color{green}{{dx}}&=\int{{\color{brown}{x}{{{\left( {\color{blue}{{3x-1}}} \right)}}^{{\frac{1}{2}}}}\color{green}{{dx}}=\int{{\left( {\color{brown}{{\frac{{u+1}}{3}}}} \right){{{\color{blue}{u}}}^{{\frac{1}{2}}}}\color{green}{{dx}}=}}}}\int{{\left( {\color{brown}{{\frac{{u+1}}{3}}}} \right){{{\color{blue}{u}}}^{{\frac{1}{2}}}}\color{green}{{\frac{{du}}{3}}}}}\\&=\frac{1}{9}\int{{\left( {{{{\color{blue}{u}}}^{{\frac{3}{2}}}}+{{{\color{blue}{u}}}^{{\frac{1}{2}}}}} \right)\color{green}{{du}}}}=\frac{1}{9}\left( {\frac{{{{{\color{blue}{u}}}^{{\frac{5}{2}}}}}}{{\frac{5}{2}}}+\frac{{{{{\color{blue}{u}}}^{{\frac{3}{2}}}}}}{{\frac{3}{2}}}} \right)+C=\frac{2}{{45}}{{\color{blue}{u}}^{{\frac{5}{2}}}}+\frac{2}{{27}}{{\color{blue}{u}}^{{\frac{3}{2}}}}+C\\&=\frac{2}{{45}}{{\left( {3x-1} \right)}^{{\frac{5}{2}}}}+\frac{2}{{27}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}+C\end{align}$

If take out the Greatest Common Factor (GCF):

$ \displaystyle =\frac{2}{{135}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}\left[ {3\left( {3x-1} \right)+5} \right]+C=\frac{2}{{135}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}\left( {9x+2} \right)+C=\frac{{2\sqrt{{{{{\left( {3x-1} \right)}}^{3}}}}\left( {9x+2} \right)}}{{135}}+C$

(Notice that $ 135$ is the Least Common Multiple (LCM) of $ 45$ and $ 27$)

Differentiate back to verify:

$ \require {cancel} \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{2}{{45}}{{{\left( {3x-1} \right)}}^{{\frac{5}{2}}}}+\frac{2}{{27}}{{{\left( {3x-1} \right)}}^{{\frac{3}{2}}}}} \right)&=\frac{{{}^{1}\cancel{5}}}{{\cancel{2}}}\cdot \frac{{\cancel{2}}}{{{{{\cancel{{45}}}}_{9}}}}{{\left( {3x-1} \right)}^{{\frac{3}{2}}}}\cdot 3+\frac{{\cancel{2}}}{{{}_{9}\cancel{{27}}}}\cdot \frac{{{{{\cancel{3}}}^{1}}}}{{\cancel{2}}}{{\left( {3x-1} \right)}^{{\frac{1}{2}}}}\cdot 3\\&=\frac{1}{3}\left( {{{{\left( {3x-1} \right)}}^{{\frac{3}{2}}}}+{{{\left( {3x-1} \right)}}^{{\frac{1}{2}}}}} \right)=\frac{1}{3}{{\left( {3x-1} \right)}^{{\frac{1}{2}}}}\left( {3x-1+1} \right)\\&={{\left( {3x-1} \right)}^{{\frac{1}{2}}}}\cdot \frac{1}{{\cancel{3}}}\left( {\cancel{3}x} \right)=x\sqrt{{3x-1}}\end{align}$ √

Which Method to Use?

What’s tricky in Calculus around now is that you’ll be expected to figure out which method of integration to use; for example, multiplication and separation of terms, $ u$-sub, and so on. On a test, they will rarely tell you to us a specific method to integrate, so you’ll have to be able to quickly identify patterns.

The following problems appear to be $ u$-sub, but actually aren’t. I like to try other methods first before trying $ u$-sub:

Integral	Setup and Solution
$ \displaystyle \int{{\frac{{4-3{{x}^{{\frac{5}{2}}}}}}{{\sqrt{x}}}}}\,dx$	$ u=?$ It looks $ u$-sub should be used here, but there’s not a complicated expression that is either raised to a power or the argument of a trig function. Just simplify the fraction and integrate: $ \displaystyle \begin{align}\int{{\frac{{4-3{{x}^{{\frac{5}{2}}}}}}{{\sqrt{x}}}}}\,dx&=\int{{\frac{4}{{\sqrt{x}}}}}-\frac{{3{{x}^{{\frac{5}{2}}}}}}{{\sqrt{x}}}\,dx=\int{{4{{x}^{{-\frac{1}{2}}}}}}-3{{x}^{{\frac{5}{2}}}}{{x}^{{-\frac{1}{2}}}}\,dx\\&=\int{{4{{x}^{{-\frac{1}{2}}}}}}-3{{x}^{2}}\,dx=8{{x}^{{\frac{1}{2}}}}-{{x}^{3}}+\,C=8\sqrt{x}-{{x}^{3}}+\,C\end{align}$ Differentiate back to verify: $ \displaystyle \frac{d}{{dx}}\left( {8{{x}^{{\frac{1}{2}}}}-{{x}^{3}}} \right)=8\cdot \frac{1}{2}\cdot {{x}^{{-\frac{1}{2}}}}-3{{x}^{2}}=4{{x}^{{-\frac{1}{2}}}}-3{{x}^{2}}=\frac{4}{{\sqrt{x}}}-3{{x}^{2}}=\frac{{4-3{{x}^{{\frac{5}{2}}}}}}{{\sqrt{x}}}$ √
$ \displaystyle \int{{{{x}^{2}}\left( {{{x}^{2}}-2x+1} \right)}}\,dx$	$ u=?$ Two reasons why we don’t want to use $ u$-sub here: 1) There isn’t a complicated expression raised to a power, and 2) If we were to take the derivative of $ \displaystyle \left( {{{x}^{2}}-2x+1} \right)$, we don’t get anything that’s outside this expression. “Push through” (distribute) the $ {{x}^{2}}$ and integrate without $ u$-sub: $ \displaystyle \begin{align}\int{{{{x}^{2}}\left( {{{x}^{2}}-2x+1} \right)}}\,dx&=\int{{{{x}^{4}}-2{{x}^{3}}+{{x}^{2}}\,dx}}=\frac{1}{5}{{x}^{5}}-\frac{2}{4}{{x}^{4}}+\frac{1}{3}{{x}^{3}}\,+\,C\\&=\frac{{{{x}^{5}}}}{5}-\frac{{{{x}^{4}}}}{2}+\frac{{{{x}^{3}}}}{3}\,+\,C\end{align}$ Differentiate back to verify: $ \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{{{{x}^{5}}}}{5}-\frac{{{{x}^{4}}}}{2}+\frac{{{{x}^{3}}}}{3}} \right)&=\frac{1}{5}\cdot 5{{x}^{4}}-\frac{1}{2}\cdot 4{{x}^{3}}+\frac{1}{3}\cdot 3{{x}^{2}}\\&={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}={{x}^{2}}\left( {{{x}^{2}}-2x+1} \right)\end{align}$ √
$ \displaystyle \int{{{{{\left( {{{x}^{2}}-3} \right)}}^{2}}}}dx$	$ u=?$ There is an expression raised to a power, but again, there is nothing outside of that expression that looks like the derivative of what’s inside. To integrate, just “FOIL” or multiply out and integrate without $ u$-sub: $ \displaystyle \begin{align}\int{{{{{\left( {{{x}^{2}}-3} \right)}}^{2}}}}\,dx&=\int{{\left( {{{x}^{4}}-6{{x}^{2}}+9} \right)\,dx}}=\frac{1}{5}{{x}^{5}}-\frac{6}{3}{{x}^{3}}+9x\,+\,C\\&=\frac{{{{x}^{5}}}}{5}-2{{x}^{3}}+9x\,+\,C\end{align}$ Differentiate back to verify: $ \displaystyle \frac{d}{{dx}}\left( {\frac{{{{x}^{5}}}}{5}-2{{x}^{3}}+9x} \right)=\frac{1}{5}\cdot 5{{x}^{4}}-2\cdot 3{{x}^{2}}+9={{x}^{4}}-6{{x}^{2}}+9={{\left( {{{x}^{2}}-3} \right)}^{2}}$ √

Don’t get discouraged; these take practice!

Understand these problems, and practice, practice, practice!

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On to Differential Equations and Slope Fields – you’re ready!