Matheno - Learn Well and Excel

u-Substitution

Straightforward summary of how to do u-substitution to evaluate an integral, along with free practice problems, each with a complete solution a click away.

Summary: u-Substitution
Hide/Show Overview

I. u-Substitution in Indefinite Integrals

If you were asked to evaluate the integral $\int \! e^{\boxdot} \, d\boxdot,$

you would probably guess that the answer is
$$\int \! e^{\boxdot} \, d\boxdot = e^{\boxdot} + C$$
no matter what exactly is in the $\boxdot$ box. You would be correct!




Similarly, these integrals are all correct:
\begin{align*}
\int \! \cos \left(\boxdot \right) \, d\boxdot &= \sin \left(\boxdot \right) + C \\[8px]
\int \! \sin \left(\boxdot \right) \, d\boxdot &= -\cos \left(\boxdot \right) + C \\[8px]
\int \! \left(\boxdot \right)^2 \, d\boxdot &= \dfrac{1}{3}\left(\boxdot \right)^3 + C
\end{align*}
What’s crucial in each instance above is that the $”\boxdot”$ is identical in both the function you’re integrating ($e^{\boxdot}$, or $\cos \left(\boxdot \right)$, …) and the $d\boxdot.$

For instance, this is correct:
$$\int \! e^{ \bbox[yellow,5px]{5x}} \, d( \bbox[yellow,5px]{5x}) = e^{ \bbox[yellow,5px]{5x}} + C$$
By constrast, if we have $d(x)$ instead of $d(5x),$ then we don’t immediately know the integral:
$$ \int \! e^{ \bbox[yellow,5px]{5x}} \, d\bbox[orange,5px]{x} \ne e^{ \bbox[yellow,5px]{5x}} + C$$
We can check that the result on the right-hand side of the equation isn’t correct because if we take its derivative, the Chain rule gives us an extra factor of 5, and so we we don’t get back the integrand:
$$\dfrac{d}{dx}\left(e^{5x} + C \right) = e^{5x} \cdot 5 + 0 \ne e^{5x} $$
We can, however, turn the integral $\int \! \,e^{5x} dx$ into one we can evaluate easily by making what’s known as a u-substitution. The process essentially consists of guessing—yes, guessing—what would be a useful variable to integrate with respect to, and then convert the integral you have into one that’s entirely in terms of that variable.



That probably sounds abstract, we know. As usual, it’s easiest (and best) to show you how it works by working through a few examples, and then you can work through many problems to try it out for yourself. We promise that with just a little practice, you’ll get good at turning the integrals you’re given into ones you already know how to evaluate.



The following example illustrates.


Example 1

Find $\int \! e^{5x} \, dx.$



Solution.

You might think, “I know that $\int \! e^u \, du = e^u + C,$” so let’s try $u = 5x$ and see what happens.
$$u = 5x$$
Then
\begin{align*}
du &= 5 \, dx \\[8px]
\implies dx &= \dfrac{1}{5}du
\end{align*}
Now let’s make those substitutions into the integral:
\begin{align*}
\int \! e^{(\overbrace{5x}^u)} \, \overbrace{dx}^{\frac{1}{5}du} &= \int \! e^u \, \left(\dfrac{1}{5} du \right) \\[8px]
&= \dfrac{1}{5} \int \! e^{\bbox[yellow,5px]{u}} \, d\bbox[yellow,5px]{u} \\[8px]
\end{align*}
$$\text{Ah, magic: we know that integral!}$$
\begin{align*}
\phantom{\int \! e^{(\overbrace{5x}^u)} \, \overbrace{dx}^{\frac{1}{5}du} } &= \dfrac{1}{5}e^{\bbox[yellow,5px]{u}} + C \\[8px]
&= \dfrac{1}{5}e^{5x} + C \quad \cmark
\end{align*}
Notice that in the last line we merely substituted back $5x = u:$ since the original problem was in terms of $x,$ our final answer also needs to be in terms of $x$ instead of $u.$



Let’s check that our answer is correct:
\begin{align*}
\dfrac{d}{dx}\left(\dfrac{1}{5}e^{5x} + C \right) &= \dfrac{1}{5}\dfrac{d}{dx}\left(e^{5x} \right) + 0 \\[8px]
&= \dfrac{1}{5}\left(e^{5x}\cdot 5 \right) \\[8px]
&= e^{5x} \quad \cmark
\end{align*}
FAQ

Why doesn’t $\dfrac{1}{5}\int \! e^u \, du = \dfrac{1}{5}\left(e^u + C \right)$? That is, why is your answer “$+C$” instead of “$+\dfrac{1}{5}C$”?

Answer: The constant $+C$ is a placeholder for some constant, and we don’t know or care what it is. By convention, we thus always write $+C$ rather than $+\dfrac{1}{5}C,$ or $-C,$ or $2C,$ or anything else.

End Example 1.


Let’s consider another example.

 


Example 2.

Find $\int \! \cos(x^2) \,x \, dx.$


Solution.

You might think, “I know that $\int \! \cos(u) \, du = \sin (u) + C,$ so let’s try $u = x^2$ and see what happens.”

If you were thinking something similar, you’re on the right track!



That is, let
$$u = x^2$$
Then
\begin{align*}
du &= 2x \, dx \\[8px]
\implies x \, dx &= \dfrac{1}{2}du
\end{align*}
Note that it’s fortunate that the original integrand has that “extra” $x$ in it: we need that $x$ in order to make the substitution $x \, dx = \dfrac{1}{2}du.$ In fact if that $x$ weren’t there, we’d be stuck and couldn’t proceed. (But then you wouldn’t be given the integral $\int \! \cos(x^2) \, dx,$ because we can’t solve it with the tools we have. We need that “extra” $x$ there.)



Let’s make the substitution $u = x^2$ into our original integral and see what happens:
\begin{align*}
\int \! \cos(\overbrace{x^2}^u) \,(\overbrace{x \, dx}^{\frac{1}{2}du}) & = \int \! \cos (u) \left( \dfrac{1}{2} du\right) \\[8px]
&= \dfrac{1}{2} \int \! \cos(\bbox[yellow,5px]{u}) \, d\bbox[yellow,5px]{u} \\[8px]
\end{align*}
$$\text{Ah, again we know that integral: }$$
\begin{align*}
\phantom{ \int \! \cos(\overbrace{x^2}^u) \,(\overbrace{x \, dx}^{\frac{1}{2}du})} &= \dfrac{1}{2}\sin(\bbox[yellow,5px]{u}) + C \\[8px]
&= \dfrac{1}{2} \sin \left(x^2 \right) + C \quad \cmark
\end{align*}
Again in the last step we substituted for $u$ in terms of $x\: \left(u = x^2 \right)$ so our final answer is in terms of $x$ instead of $u.$



Let’s check that our answer is correct:
\begin{align*}
\dfrac{d}{dx}\left(\dfrac{1}{2} \sin \left(x^2 \right) + C \right) &= \dfrac{1}{2} \dfrac{d}{dx}\sin(x^2) + 0 \\[8px]
&= \dfrac{1}{2}\cos(x^2)\cdot \dfrac{d}{dx}\left(x^2 \right) \\[8px]
&= \dfrac{1}{2}\cos(x^2)\cdot (2x) \\[8px]
&= \cos(x^2) \: x \quad \cmark
\end{align*}
End Example 2.


The upshot: As you can see, making a u-substitution can quickly turn an integral you don’t immediately know into one that you do. To do so, guess what a good choice for $u$ is, and then see what happens.


 

II. u-Substitution in Definite Integrals

If you’re given a definite integral (with limits of integration), then it’s easiest to convert those $x$ values into their equivalent $u$ values and then complete the calculation in terms of $u$. The following example illustrates.

Example 3.

Find $\int_0^1 \! e^{5x} \, dx.$




Solution.

As in Example 1 above, let $u = 5x:$
\begin{align*}
u &= 5x \\[8px]
du &= 5 \, dx \\[8px]
\implies dx &= \dfrac{1}{5}du
\end{align*}
We must also convert the limits of integration to be in terms of $u:$
\begin{align*}
u &= 5x \\[8px]
\text{When }x=0: \quad u &= 5(0) = 0 \\[8px]
\text{When }x=1: \quad u &= 5(1) = 5
\end{align*}
Now let’s make those substitutions into the integral, changing the limits of integration too:
\begin{align*}
\int_0^1 \! e^{(\overbrace{5x}^u)} \, \overbrace{dx}^{\frac{1}{5}du} &= \int_\bbox[yellow,5px]{0}^\bbox[yellow,5px]{5} \! e^u \, \left(\dfrac{1}{5} du \right) \\[8px]
&= \dfrac{1}{5} \int_0^5 \! e^{u} \, du \\[8px]
&= \dfrac{1}{5}\left[ e^{u}\right]_0^5 \\[8px]
&= \dfrac{1}{5}\left[e^5 – e^0 \right] \\[8px]
&= \dfrac{1}{5}\left[e^5 – 1 \right] \quad \cmark
\end{align*}
End Example 3.


[collapse]

There are many more example problems below so you can get the hang of how to do u-substititions.

Problem #1
(a) Find $\int \! (5x + 27)^{98} \, dx.$
(b) Find $\int_0^1 \! (5x + 27)^{98} \, dx.$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) $\dfrac{1}{495} (5x + 27)^{99} + C$
(b) $\dfrac{1}{495} \left[(32)^{99} – (27)^{99} \right]$

Let $u = 5x + 27:$ \begin{align*} u &= 5x + 27 \\[8px] du &= 5\, dx \implies dx = \dfrac{1}{5}du \end{align*} Then \begin{align*} \int \! (\overbrace{5x + 27 }^u)^{98} \overbrace{\, dx}^{\frac{1}{5}du} &= \int \! u^{98} \, \left(\dfrac{1}{5}du \right) \\[8px] &= \dfrac{1}{5} \int \! u^{98} \, du \\[8px] &= \dfrac{1}{5} \cdot \dfrac{1}{99}u^{99} + C \\[8px] &= \dfrac{1}{495} (5x + 27)^{99} + C \quad \cmark \end{align*}
Let $u = 5x + 27:$ \begin{align*} u &= 5x + 27 \\[8px] du &= 5\, dx \implies dx = \dfrac{1}{5}du \end{align*} Change limits: \begin{align*} u &= 5x + 27 \\[8px] \text{When }x=0: \quad u &= 5(0) + 27 = 27 \\[8px] \text{When }x=1: \quad u &= 5(1) + 27 = 32 \end{align*} Then \begin{align*} \int_0^1 \! (\overbrace{5x + 27 }^u)^{98} \overbrace{\, dx}^{\frac{1}{5}du} &= \int_{27}^{32} \! u^{98} \, \left(\dfrac{1}{5}du \right) \\[8px] &= \dfrac{1}{5} \int_{27}^{32} \! u^{98} \, du \\[8px] &= \dfrac{1}{5} \cdot \dfrac{1}{99}\left[ u^{99}\right]_{27}^{32} \\[8px] &= \dfrac{1}{495} \left[(32)^{99} – (27)^{99} \right] \quad \cmark \end{align*}
[hide solution]
Problem #2
Find $\int \! \sin(2x -2) \, dx.$
Show/Hide Solution
Let $u = 2x -2:$ \begin{align*} u &= 2x – 2 \\[8px] du &= 2 dx \implies dx = \dfrac{1}{2}du \end{align*} Then \begin{align*} \int \! \sin(\overbrace{2x -2}^u) \overbrace{ \, dx}^{\frac{1}{2}du} &= \int \! \sin(u) \, \left(\dfrac{1}{2}du \right) \\[8px] &= \dfrac{1}{2} \int \! \sin(u) \, du \\[8px] &= -\dfrac{1}{2}\cos(u) + C \\[8px] &= -\dfrac{1}{2}\cos(2x -2) + C \quad \cmark \end{align*}
[hide solution]
Problem #3
(a) Find $\int \! \sqrt{3x + 2} \, dx.$
(b) Find $\int_1^3 \! \sqrt{3x + 2} \, dx.$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) $\dfrac{2}{9} (3x + 2)^{3/2} + C$
(b) $\dfrac{2}{9} \left[11 \sqrt{11} – 5 \sqrt{5} \right]$

Let $u = 3x + 2:$ \begin{align*} u &= 3x + 2 \\[8px] du &= 3 \, dx \implies dx = \dfrac{1}{3}du \end{align*} Then: \begin{align*} \int \! \overbrace{\sqrt{3x + 2}}^\sqrt{u} \, \overbrace{dx}^{\frac{1}{3}du} &= \int \! \sqrt{u} \, \left( \dfrac{1}{3}du\right) \\[8px] &= \dfrac{1}{3} \int \! \sqrt{u} \, du \\[8px] &= \dfrac{1}{3} \cdot \dfrac{1}{\frac{3}{2}} u^{3/2} + C\\[8px] &= \dfrac{1}{3} \cdot \dfrac{2}{3} u^{3/2} + C \\[8px] &= \dfrac{2}{9} u^{3/2} + C \\[8px] &= \dfrac{2}{9} (3x + 2)^{3/2} + C \quad \cmark \end{align*}
Let $u = 3x + 2:$ \begin{align*} u &= 3x + 2 \\[8px] du &= 3 \, dx \implies dx = \dfrac{1}{3}du \end{align*} Change limits: \begin{align*} u &= 3x + 2 \\[8px] \text{When }x=1: \quad u &= 3(1) + 2 = 5 \\[8px] \text{When }x=3: \quad u &= 3(3) + 2 = 11 \end{align*} Then: \begin{align*} \int_1^3 \! \overbrace{\sqrt{3x + 2}}^\sqrt{u} \, \overbrace{dx}^{\frac{1}{3}du} &= \int_5^{11} \! \sqrt{u} \, \left( \dfrac{1}{3}du\right) \\[8px] &= \dfrac{1}{3} \int_5^{11} \! \sqrt{u} \, du \\[8px] &= \dfrac{1}{3} \cdot \dfrac{1}{\frac{3}{2}} \left[u^{3/2} \right]_5^{11}\\[8px] &= \dfrac{1}{3} \cdot \dfrac{2}{3} \left[(11)^{3/2} – (5)^{3/2} \right] \\[8px] &= \dfrac{2}{9} \left[(11)^{3/2} – (5)^{3/2} \right] \quad \cmark \\[8px] &= \dfrac{2}{9} \left[11 \sqrt{11} – 5 \sqrt{5} \right] \quad \cmark \end{align*}
[hide solution]
Problem #4
Find $\int \! \dfrac{dx}{\sqrt[3]{6x - 5}} \, dx.$
Show/Hide Solution
Let $u = 6x – 5:$ \begin{align*} u &= 6x – 5 \\[8px] du &= 6\, dx \implies dx = \dfrac{1}{6}du \end{align*} Then \begin{align*} \int \! \dfrac{1}{\sqrt[3]{6x – 5}}\, dx &= \int \!\dfrac{1}{\sqrt[3]{u}} \, \left(\dfrac{1}{6}du \right) \\[8px] &= \dfrac{1}{6}\int \! u^{-3/2} \, du \\[8px] &= \dfrac{1}{6} \cdot \dfrac{1}{-\frac{1}{2}} u^{-1/2} + C \\[8px] &= \dfrac{1}{6} \cdot (-2)u^{-1/2} + C \\[8px] &= -\dfrac{1}{3} u^{-1/2} + C \\[8px] &= -\dfrac{1}{3} (6x -5)^{-1/2} + C \quad \cmark \\[8px] &= -\dfrac{1}{3} \dfrac{1}{\sqrt{6x -5}} + C \quad \cmark \end{align*}
[hide solution]
Problem #5
Find $\int \! \sec^2\left(\dfrac{x}{5} \right) \, dx.$
Show/Hide Solution
Let $u = \dfrac{x}{5}:$ \begin{align*} u &= \dfrac{x}{5} \\[8px] du &= \dfrac{1}{5}dx \implies dx = 5\, du \end{align*} Then \begin{align*} \int \! \sec^2\overbrace{\left(\dfrac{x}{5} \right)}^u \overbrace{\, dx}^{5\,du} &= \int \! \sec^2(u) \, (5\, du) \\[8px] &= 5\int \! \sec^2(u) \,du \\[8px] &= 5\tan(u) + C \\[8px] &= 5\tan\left( \dfrac{x}{5}\right) + C \quad \cmark \end{align*}
[hide solution]

Buy us a coffee We're working to add more,
and would appreciate your help
to keep going! 😊

Problem #6
Find $\int \!x\left(x^2+ 3\right)^9 \, dx.$
Show/Hide Solution
Let $u = x^2 + 3:$ \begin{align*} u &= x^2 + 3 \\[8px] du &= 2x \, dx \implies x \,dx = \dfrac{1}{2}du \end{align*} Then \begin{align*} \int \!(\overbrace{x^2+ 3}^u)^9 \overbrace{\, (x\,dx)}^{\frac{1}{2}du} &= \int \! u^9 \, \left(\dfrac{1}{2}du \right) \\[8px] &= \dfrac{1}{2}\int \! u^9 \, du \\[8px] &= \dfrac{1}{2}\cdot \dfrac{1}{10} u^{10} + C \\[8px] &= \dfrac{1}{20}(x^2 + 3)^{10} + C \quad \cmark \end{align*}
[hide solution]
Problem #7
Find $\int \! (x^3 - 2x^2)^{5}\left(3x^2 - 4x \right) \, dx.$
Show/Hide Solution
Let $u = x^3 – 2x^2:$ \begin{align*} u &= x^3 – 2x^2 \\[8px] du &= \left(3x^2 -4x \right)dx \end{align*} Then \begin{align*} \int \! (\overbrace{x^3 – 2x^2}^u)^{5}\,\overbrace{\left(3x^2 – 4x \right) dx}^{du} &= \int \! u^5 \, du \\[8px] &= \dfrac{1}{6}u^6 + C \\[8px] &= \dfrac{1}{6}\left( x^3 – 2x^2\right)^6 + C \quad \cmark \end{align*}
[hide solution]
Problem #8
Find $\int \!\cos \left(e^x \right) \, e^x \, dx.$
Show/Hide Solution
Let $u = e^x:$ \begin{align*} u &= e^x \\[8px] du &= e^x dx \end{align*} Then \begin{align*} \int \!\cos \overbrace{\left( e^x\right)}^u \, \overbrace{\left( e^x \, dx\right)}^{du} &= \int \cos(u) \, du \\[8px] &= \sin(u) + C \\[8px] &= \sin \left(e^x \right) + C \quad \cmark \end{align*}
[hide solution]
Problem #9
Find $\int \! \sec \left(e^x \right) \tan \left( e^x\right)\, e^x\, dx.$
Show/Hide Solution
Let $u = e^x:$ \begin{align*} u &= e^x \\[8px] du &= e^x \, dx \end{align*} Then: \begin{align*} \int \! \sec \overbrace{\left(e^x \right)}^u \tan \overbrace{\left( e^x\right)}^u \overbrace{\left(e^x\, dx \right)}^{du} &= \int \! \sec(u) \tan(u)\, du \\[8px] &= \sec(u) + C \\[8px] &= \sec \left( e^x\right) + C \quad \cmark \end{align*}
[hide solution]
Problem #10
Find $\int \! \dfrac{\cos \sqrt{x}}{\sqrt{x}} \, dx.$
Show/Hide Solution
Let $u = \sqrt{x}:$ \begin{align*} u &= \sqrt{x} \\[8px] du &= \dfrac{1}{2}\dfrac{1}{\sqrt{x}}dx \implies \dfrac{dx}{\sqrt{x}} = 2\,du \end{align*} Then \begin{align*} \int \!\cos \overbrace{\sqrt{x}}^u \, \overbrace{\left(\dfrac{dx}{\sqrt{x}}\right)}^{2 \, du} &= \int \! \cos u \, (2\,du) \\[8px] &= 2 \int \! \cos u \, du \\[8px] &= 2 \sin u + C \\[8px] &= 2 \sin \sqrt{x} + C \quad \cmark \end{align*}
[hide solution]
Problem #10
Find $\int \! \sin x \cos x \, dx.$
Show/Hide Solution
Method 1.
Let $u = \sin x:$ \begin{align*} u &= \sin x \\[8px] du &= \cos x dx \end{align*} Then \begin{align*} \int \! \overbrace{\sin x}^u (\overbrace{\cos x \, dx}^{du}) &= \int \! u \, du \\[8px] &= \dfrac{1}{2}u^2 + C \\[8px] &= \dfrac{1}{2}\sin^2 x + C \quad \cmark \end{align*}
Method 2.
Let $u = \cos x:$ \begin{align*} u &= \cos x \\[8px] du &= -\sin x \, dx \implies \sin x \, dx = -du \end{align*} Then \begin{align*} \int \! \sin x \cos x \, dx &= \int \! \overbrace{\cos x}^u (\overbrace{\sin x \, dx}^{-du})\\[8px] &= \int \! u\, (-du) \\[8px] &= -\int\! u \, du \\[8px] &= – \dfrac{1}{2}u^2 + C^* \\[8px] &= -\dfrac{1}{2}\cos^2 x + C^* \quad \cmark \end{align*} FAQ.
How can the answers you get using Method 1 ($\frac{1}{2}\sin^2 x + C$) and Method 2 ($-\frac{1}{2}\cos^2 x + C^*$) both be right??
Answer: To see that the two answers actually are equivalent, remember the trig identity $\sin^2 x + \cos^2 x = 1.$ That means \[\sin^2 x = 1 – \cos^2 x \] Then let’s start with the answer we got using Method 1: \begin{align*} \int \! \sin x \cos x \, dx &= \dfrac{1}{2}\sin^2 x + C \\[8px] &= \dfrac{1}{2}\left( 1 – \cos^2 x\right) + C \\[8px] &= -\dfrac{1}{2}\cos^2 x + \dfrac{1}{2} + C \end{align*} Compare that to the answer we got using Method 2: \[\int \! \sin x \cos x \, dx = -\dfrac{1}{2}\cos^2 x + C^*\] We then see that the two answers are the same; the constants $C$ and $C^*$ are just different (and in fact if we care: $C^* = \dfrac{1}{2} + C$). Either one is still just a consant.
[hide solution]
Problem #11
Find $\int \! \sin^2 x \cos x \, dx.$
Show/Hide Solution
Let $u = \sin x:$ \begin{align*} u & = \sin x \\[8px] du &= \cos x \, dx \end{align*} Then \begin{align*} \int \! \sin^2 x \cos x \, dx &= \int \! (\overbrace{\sin x}^u)^2 (\overbrace{\cos x \, dx}^{du}) \\[8px] &= \int \! u^2 \, du \\[8px] &= \dfrac{1}{3}u^3 + C \\[8px] &= \dfrac{1}{3} \sin^3 x + C \quad \cmark \end{align*}
[hide solution]
Problem #12
Find $\int \! \tan x \sec^2 x \, dx.$
Show/Hide Solution
Let $u = \tan x:$ \begin{align*} u &= \tan x \\[8px] du &= \sec^2 x \, dx \end{align*} Then \begin{align*} \int \! \overbrace{\tan x}^u (\overbrace{\sec^2 x \, dx}^{du}) &= \int \! u \, du \\[8px] &= \dfrac{1}{2}u^2 + C \\[8px] &= \dfrac{1}{2}\tan^2 x + C \quad \cmark \end{align*}
[hide solution]
Problem #13
Find $\int \! \sqrt{\cot x} \csc^2 x \, dx.$
Show/Hide Solution
Let $u = \cot x:$ \begin{align*} u &= \cot x \\[8px] du &= -\csc x \, dx \implies \csc x \, dx = -du \end{align*} Then \begin{align*} \int \! \overbrace{\sqrt{\cot x}}^{\sqrt{u}} (\overbrace{\csc^2 x \, dx}^{-du}) &= \int \! \sqrt{u} \, (-du) \\[8px] &= – \int \! \sqrt{u} \, du \\[8px] &= – \dfrac{1}{\frac{3}{2}}u^{3/2}+ C \\[8px] &= -\dfrac{2}{3}(\cot x)^{3/2} + C \quad \cmark \end{align*}
[hide solution]
Problem #14
(a) Find $\int \! \dfrac{dx}{x (\ln x)^2 } \, dx.$
(b) Evaluate $\int_e^{e^2} \! \dfrac{dx}{x (\ln x)^2 } \, dx.$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) $-\dfrac{1}{2}\dfrac{1}{(\ln x)^2} + C$
(b) $\dfrac{3}{8}$

Let $u = \ln x:$ \begin{align*} u &= \ln x \\[8px] du &= \dfrac{dx}{x} \end{align*} Then \begin{align*} \int \! \dfrac{1}{(\ln x)^3}\dfrac{dx}{x} &= \int \! \dfrac{1}{u^3} \, du \\[8px] &= \int \! u^{-3} \, du \\[8px] &= \dfrac{1}{(-2)}u^{-2} + C \\[8px] &= -\dfrac{1}{2}(\ln x)^{-2} + C \\[8px] &= -\dfrac{1}{2}\dfrac{1}{(\ln x)^2} + C \quad \cmark \end{align*}
Let $u = \ln x:$ \begin{align*} u &= \ln x \\[8px] du &= \dfrac{dx}{x} \end{align*} Change limits: \begin{align*} u &= ln x \\[8px] \text{When }x=e: \quad u &= \ln e = 1 \\[8px] \text{When }x=e^2: \quad u &= \ln e^2 = 2 \ln e = 2 \\[8px] \end{align*} Then \begin{align*} \int_e^{e^2} \! \dfrac{1}{(\ln x)^3}\dfrac{dx}{x} &= \int_1^2 \! \dfrac{1}{u^3} \, du \\[8px] &= \int_1^2 \! u^{-3} \, du \\[8px] &= \left[\dfrac{1}{(-2)}u^{-2} \right]_1^2 \\[8px] &= -\dfrac{1}{2}\left[\dfrac{1}{u^2} \right]_1^2 \\[8px] &= -\dfrac{1}{2} \left[\dfrac{1}{4} – 1 \right] \\[8px] &= -\dfrac{1}{2} \left[-\dfrac{3}{4} \right] \\[8px] &= \dfrac{3}{8} \quad \cmark \end{align*}
[hide solution]
Problem #15
Find $\int \! \csc(\pi x) \cot(\pi x) \, dx.$
Show/Hide Solution
Remember that $\dfrac{d}{dx}\csc u = – \csc u \cot u.$ Hence let $u = \pi x:$ \begin{align*} u &= \pi x \\[8px] du &= \pi dx \implies dx = \dfrac{1}{\pi}du \end{align*} Then \begin{align*} \int \! \csc(\pi x) \cot(\pi x) \, dx &= \int \! \csc(u) \cot(u) \, \left(\dfrac{1}{\pi}du \right) \\[8px] &= \dfrac{1}{\pi} \int \! \csc(u) \cot(u) \, du \\[8px] &= \dfrac{1}{\pi} (-\csc(u)) + C \\[8px] &= -\dfrac{1}{\pi} \csc(\pi x) + C \quad \cmark \end{align*}
[hide solution]
Problem #16
Evaluate $\int_0^{\pi/4} \! e^{\tan x} \sec^2 x \, dx.$
Show/Hide Solution
Ideally you recognize $\sec^2 x$ as the derivative of $\tan x.$ That suggests that we should let $u = \tan x:$ \begin{align*} u &= \tan x \\[8px] du &= \sec^2 x \, dx \end{align*} Change limits: \begin{align*} u &= \tan x \\[8px] \text{When }x=0: \quad u &= \tan 0 = 0 \\[8px] \text{When }x=\dfrac{\pi}{4}: \quad u &= \tan \dfrac{\pi}{4} = 1 \\[8px] \end{align*} Then: \begin{align*} \int_0^{\pi/4} \! e^{\overbrace{\tan x}^u} \overbrace{(\sec^2 x \, dx)}^{du} &= \int_0^1 \! e^u \, du \\[8px] &= \left[e^u \right]_0^1 \\[8px] &= e^1 – e^0 \\[8px] &= e – 1 \quad \cmark \end{align*}
[hide solution]
Problem #17
Evaluate $\int_0^1 \! \dfrac{e^x + 2x }{e^x + x^2} \, dx.$
Show/Hide Solution
You might recognize the numerator $\left(e^x + 2x \right)$ as the derivative of the denominator $\left(e^x + x^2 \right).$ That’s a good clue that we should make $u$ equal to the denominator: \begin{align*} u &= e^x + x^2 \\[8px] du &= (e^x + 2x) dx \end{align*} Change limits: \begin{align*} u &= e^x + x^2 \\[8px] \text{When }x=0: \quad u &= e^0 + 0 = 1 \\[8px] \text{When }x=1: \quad u &= e^1 + 1 = e + 1 \\[8px] \end{align*} Then: \begin{align*} \int_0^1 \! \dfrac{e^x + 2x }{e^x + x^2} \, dx &= \int_{1}^{e+1} \! \dfrac{du}{u} \\[8px] &= \left[ \ln(u)\right]_{1}^{e+1} \\[8px] &= \ln(e+1) – \cancelto{0}{\ln(1)} \\[8px] &= \ln(e+1) \quad \cmark \end{align*}
[hide solution]
Less clear u-substitutions
The first u-substitution problems you'll encounter will probably be like the ones above, where (with practice) you'll come to recognize what u should be to turn the integral into one you know how to evaluate. For example, all of the ones above where you end up with something like $\int \! e^u \, du,$ $\int \! \cos(u) \, du,$ and so forth. In other problems, though, you'll look at the integral and think, "I don't recognize what to do here." That thought itself is a clue that you should try a u-substitution. Again, you have to just guess what u is, and then proceed and see what happens; if one approach doesn't work, make a different guess for what u is and then try again. The following problems illustrate.
Problem #18
Find $\int \! x\sqrt{x - 3} \, dx.$
Show/Hide Solution
We don’t immediately see how to evaluate this integral — but a helpful move is often to take what’s under the square-root sign $(x-3)$ and turn it into our new variable $u.$ So let’s try $u = x-3:$ \begin{align*} u &= x -3 \implies x = u + 3 \\[8px] du&= dx \end{align*} Now we’ll make our substitutions, including $x = u + 3:$ \begin{align*} \int \! \overbrace{x}^{u+3} \overbrace{\sqrt{x – 3}}^{\sqrt{u}} \, \overbrace{dx}^{du} &= \int \! (u+3)\sqrt{u} \, du \\[8px] \end{align*} $$\text{Ah, this is looking promising:}$$ \begin{align*} \phantom{\int \! \overbrace{x}^{u+3} \overbrace{\sqrt{x – 3}}^{\sqrt{u}} \, \overbrace{dx}^{du}} &= \int \! \left(u^{3/2} + 3 \sqrt{u} \right) \, du \\[8px] &= \int \! u^{3/2} \, du + 3 \int \! u^{1/2} \, du \\[8px] &= \dfrac{1}{\frac{5}{2}}u^{5/2} + 3 \cdot \dfrac{1}{\frac{3}{2}}u^{3/2} + C \\[8px] &= \dfrac{2}{5}u^{5/2} + 2 u^{3/2} + C \\[8px] &= \dfrac{2}{5}(x-3)^{5/2} + 2(x-3)^{3/2} + C \quad \cmark \end{align*}
[hide solution]
Problem #19
Find $\int \! \dfrac{x}{\sqrt{x + 5}} \, dx.$
Show/Hide Solution
Let’s again try making $u$ equal to what’s under the square-root sign, $u = x + 5:$ \begin{align*} u &= x + 5 \implies x = u -5 \\[8px] du &= dx \end{align*} Then \begin{align*} \int \! \dfrac{x}{\sqrt{x + 5}} \, dx &= \int \! \dfrac{u-5}{\sqrt{u}} \, du \\[8px] &= \int \! \left(\sqrt{u} – 5 u^{-1/2} \right) \, du \\[8px] &= \dfrac{1}{\frac{3}{2}}u^{3/2} – 5 \cdot \dfrac{1}{\frac{1}{2}}u^{1/2} + C \\[8px] &= \dfrac{2}{3}u^{3/2} -10u^{1/2} + C \\[8px] &= \dfrac{2}{3}(x+5)^{3/2} – 10 \sqrt{x+5} + C \quad \cmark \end{align*}
[hide solution]
Problem #20
Find $\int \! 3\left(x^3 - 2 \right)^{1/4} x^5 \, dx.$
Show/Hide Solution
Again it’s not immediately clear what to do, so let’s set $u$ equal to what’s in the parentheses, $u = x^3 – 2.$
\begin{align*} u &= x^3 – 2 \\[8px] du &= 3x^2 \, dx \end{align*} Then \begin{align*} \int \! 3\left(x^3 – 2 \right)^{1/4} x^5 \, dx &= \int \! \overbrace{\left(x^3 – 2 \right)^{1/4}}^{u^{1/4}} x^3 \overbrace{(3x^2 \, dx)}^{du} \end{align*} Hmmm. Now we have an “extra” $x^3$ in there that we need to rewrite in terms of $u.$ To do so, note that \[u = x^3 – 2 \implies x^3 = u + 2\] Hence \begin{align*} \int \! 3\left(x^3 – 2 \right)^{1/4} x^5 \, dx &= \int \! \overbrace{\left(x^3 – 2 \right)^{1/4}}^{u^{1/4}} x^3 \overbrace{(3x^2 \, dx)}^{du} \\[8px] &= \int \!u^{1/4}(u + 2)\, du \\[8px] &= \int \!\left(u^{5/4} + 2u^{1/4} \right) \, du \\[8px] &= \dfrac{1}{\frac{9}{4}}u^{9/4} + 2 \cdot \dfrac{1}{\frac{5}{4}}u^{5/4} + C \\[8px] &= \dfrac{4}{9}(x^3 – 2)^{9/4} + \dfrac{8}{5}(x^3 – 2)^{5/4} + C \quad \cmark \end{align*}
[hide solution]
Problem #21
Find $\int \! \sqrt{1 + \sqrt{x}} \, dx.$
Show/Hide Solution
This looks tough! But yet again: sometimes you just have to take your best guess for $u$ and see what happens.
Our first guess is to let $u$ be what’s under the square-root sign: Let $u = 1 + \sqrt{x}:$ \begin{align*} u &= 1 + \sqrt{x} \\[8px] du &= \dfrac{1}{2}\dfrac{1}{\sqrt{x}}dx \implies \dfrac{1}{\sqrt{x}} dx = 2 \, du \end{align*} Hmmm. That’s not great, because we don’t have $\dfrac{1}{\sqrt{x}} \, dx$ in the original integrand.
Let’s press forward, and solve for $dx$ to see if we can make something work there: \begin{align*} &\phantom{ = \dfrac{1}{2}\dfrac{1}{\sqrt{x}}dx \implies \dfrac{1}{\sqrt{x}} dx = 2 \, du}\\ dx &= 2 \sqrt{x} \, du \end{align*} Well, we could turn that $\sqrt{x}$ into something with $u:$
Since \[ u = 1 + \sqrt{x}\] we have \[\sqrt{x} = u – 1 \] That means \begin{align*} dx &= 2 \sqrt{x} \, du \\[8px] &= 2 (u-1) \, du \end{align*} Ah: now we can turn all of our $x$’s in the original integrand into $u$’s! \begin{align*} \int \! \overbrace{\sqrt{1 + \sqrt{x}}}^{\sqrt{u}} \, \overbrace{dx}^{2(u-1)du} &= \int \!\sqrt{u} \, \big(2(u-1)\,du \big) \\[8px] &= 2\int \! (u^{3/2} – \sqrt{u}) \, du \\[8px] &= 2\left[\dfrac{1}{\frac{5}{2}}u^{5/2} – \dfrac{1}{\frac{3}{2}}u^{3/2} \right] + C \\[8px] &= 2 \left[\dfrac{2}{5}u^{5/2} – \dfrac{2}{3}u ^{3/2} \right] + C \\[8px] &= \dfrac{4}{5}\left( 1 + \sqrt{x}\right)^{5/2} + \dfrac{4}{3}\left( 1 + \sqrt{x}\right)^{3/2} + C \quad \cmark \end{align*} Let’s be clear: we were not(!) sure initially that our guess $u = 1 + \sqrt{x}$ would work when we started the solution. Really: not at all. And it might not have, in which case we would have had to try something else.
The key lesson: start somewhere, and see where it takes you!
[hide solution]
Problem #22
If $f$ is continuous and $\int_0^3 \! f(x) \, dx = 5,$ find $\int_0^1 \! f(3x) \, dx.$
Show/Hide Solution
In the integral we’re after, let $u = 3x:$ \begin{align*} u &= 3x \\[8px] du &= 3\,dx \implies dx = \dfrac{1}{3} du \end{align*} Change limits: \begin{align*} u &= 3x \\[8px] \text{When }x=0: \quad u &= 0 \\[8px] \text{When }x=1: \quad u &= 3 \end{align*} Then \begin{align*} \int_0^1 \! f(3x) \, dx &= \int_0^3 \! f(u) \, \left( \dfrac{1}{3} du\right) \\[8px] &= \dfrac{1}{3}\int_0^3 \! f(u) \, du \end{align*} Remember that the $x$ in any integral is a “dummy variable,” meaning we can call it whatever we want. So in the first integral the problem told us about, let’s call that variable $u$ instead:
$\quad \int_0^3 \! f(x) \, dx = \int_0^3 \! f(u) \, du = 5.$
Hence returning to the integral we’re after: \begin{align*} \int_0^1 \! f(3x) \, dx &= \dfrac{1}{3}\,\overbrace{\int_0^3 \! f(u) \, du}^5 \\[8px] &= \dfrac{1}{3} \cdot 5 \\[8px] &= \dfrac{5}{3} \quad \cmark \end{align*}
[hide solution]
Problem #23
If $f$ is continuous and $\int_1^{25} \! f(x) \, dx = 9,$ find $\int_1^5 \! xf(x^2) \, dx.$
Show/Hide Solution
In the integral we’re after, let $u = x^2:$ \begin{align*} u &= x^2 \\[8px] du &= 2x\, dx \implies x\,dx = \dfrac{1}{2}du \end{align*} Change limits: \begin{align*} u &= x^2 \\[8px] \text{When }x=1: \quad u &= 1 \\[8px] \text{When }x=5: \quad u &= 25 \end{align*} Then: \begin{align*} \int_1^5 \! f(x^2) \, (x\, dx) &= \int_1^{25} \! f(u) \, \left(\dfrac{1}{2}du \right) \\[8px] &= \dfrac{1}{2}\int_1^{25} \! f(u) \, du \\[8px] \end{align*} Remember that the $x$ in any integral is a “dummy variable,” meaning we can call it whatever we want. So in the first integral the problem told us about, let’s call that variable $u$ instead:
$\quad \int_1^{25} \! f(x) \, dx = \int_1^{25} \! f(u) \, du = 9.$
Hence returning to the integral we’re after: \begin{align*} \int_1^5 \! f(x^2) \, (x\, dx) &= \dfrac{1}{2}\overbrace{\int_1^{25} \! f(u) \, du}^9 \\[8px] &= \dfrac{1}{2} \cdot 9 \\[8px] &= \dfrac{9}{2} \quad \cmark \end{align*}
[hide solution]

Buy us a coffee If we've helped, please consider
giving a little something back.
Thank you! 😊


We'd love to hear:
  • What questions do you have about the solutions above?
  • Which ones are giving you the most trouble?
  • What other integration problems are you trying to work through for your class?
If you post on our Forum, we'll do our best to help!
As of September 2022, we’re using our Forum for comments and discussion of this topic, and for any math questions. We’d love to see you there! Please tap to visit our Forum: community.matheno.com.
We'd appreciate your feedback! 😊
How helpful?