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

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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.

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.

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.

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.

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

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Solution SummarySolution (a) DetailSolution (b) Detail

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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*}

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Problem #3

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Solution SummarySolution (a) DetailSolution (b) Detail

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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*}

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[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*}

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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*}

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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*}

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[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*}

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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*}

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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*}

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[hide solution]

Problem #10

Find $\int \! \sin x \cos x \, dx.$ *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.

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Show/Hide Solution

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*}

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*}

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??

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[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*}

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[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*}

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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*}

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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*}

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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*}

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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*}

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[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*}

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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*}

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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*}

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\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*}

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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!

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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

The key lesson: start

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[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*}

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$\quad \int_0^3 \! f(x) \, dx = \int_0^3 \! f(u) \, du = 5.$

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*}

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[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*}

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$\quad \int_1^{25} \! f(x) \, dx = \int_1^{25} \! f(u) \, du = 9.$

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*}

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Buy us a coffeeWe're working to add more,and would appreciate your help

to keep going! 😊