Why Substitute?

Evaluate $\displaystyle\int \cos(2x) \ dx$.

Solution

If the problem were $\displaystyle\int \cos x \ dx$, the answer would simply be $\ans{\sin x + C}$.

It's not quite that simple, but let's use that as a starting point. In other words, think about what function would have $\cos(2x)$ as its derivative - you should be able to tell that the answer will involve $\sin (2x)$.

The answer can't only be $\sin(2x) + C,$ though, because the derivative of that (using the Chain Rule) is $2 \cos (2x).$

We're getting close, though. The integral of $\cos (2x)$ is almost $\sin (2x) + C$, but we're off by a multiple of 2.

Check this out, though: since our answer is double what it should be, we know that the correct answer is half of our guess.

Putting this all together: $\int \cos (2x) \ dx = \ans{\dfrac{1}{2}\sin(2x) + C}$

Check

We can verify this by differentiating $\dfrac{1}{2}\sin(2x) + C$; I'll leave it to you to make sure its derivative is $\cos(2x)$.

First Substitution

Evaluate $\displaystyle\int 2 \cos (2x) \ dx$.

Solution

At this point, it shouldn't be huge surprise that the answer will be $\sin(2x) + C,$ but our goal is to see how u-substitution can get us to this answer.

Remember, we have a very short list of functions that we actually know how to integrate: anything of the form $x^n$, $e^x$, and a handful of trigonometric functions.

To integrate $\cos(2x)$, look for the closest form that we do know how to integrate: $\cos x$. If instead of $2x$ we had $x$, we'd be able to integrate this.

That's what leads us to substitution: we're going to introduce a new variable $u$ so that our problem just involves integrating $\cos u$. To do this, we need to let $u = 2x.$

Now our problem looks like $\int 2 \cos (2x) \ dx = \int 2 \cos u \ dx$

That raises a new issue, though: we've got a variable mismatch. We're integrating a function of $u$ with respect to $x$, which doesn't make any sense. Instead of $dx$ at the end, we need to have $du$.

Differentials

We need to take our definition of $u$ and find $du$ from it. Without getting too technical, $du = \dfrac{du}{dx} \ dx$

So since $u = 2x$, $\dfrac{du}{dx} = 2$, so $du = 2\ dx$

In every problem, once you define $u$, take its derivative, slap on $dx$, and you've got $du$.

Back to the Problem

Where were we? $\int 2 \cos (2x) \ dx = \int {\color{red}2} \cos u \ {\color{red}dx}$

Look at that: we've got a $2$ and a $dx$ in our problem, so we can replace those with ${\color{red}du}$.

$\int 2 \cos (2x) \ dx = \int {\color{red}2} \cos u \ {\color{red}dx} = \int \cos u \ du$

At this point we can integrate: $\int \cos u \ du = \sin u + C$

We've got an answer, but it is given in terms of $u$, which isn't what we want; the problem was given in terms of $x$, so the answer should be too.

All we need to do now is replace $u$ with $2x$ (look back at where we defined $u$), and we're all done!

$\int 2 \cos (2x) \ dx = \ans{\sin (2x) + C}$

Exponential u-Substitution

Use u-substitution to evaluate $\displaystyle\int 8e^{8x} \ dx$.

Solution

Let's follow the process outlined above to pick $u$. Remember that, after substituting, we want the result to be one of the basic integrals that we can do: we can integrate $e^u$, but not $e^{8x}$. Plus, we notice that $8$ is the derivative of $8x$, so if we let $u=8x$, the $8$ will get picked up by the $du$. Let's try that: $u=8x \longrightarrow du = 8 \ dx$

Now make the substitution: $\int e^u \ du$

We can integrate this, and once we do, we can substitute back into $x$ to get the final answer: $\int e^u \ du = e^u + C = \ans{e^{8x} + C}$

Check

As always, we can check by differentiating our answer: $\dfrac{d}{dx} (e^{8x} + C) = 8e^{8x} \textrm{(using the Chain Rule)}$

Power Function u-Substitution

Evaluate $\displaystyle\int (2x^3-3)^4 \cdot 6x^2\ dx$.

Solution

The basic form here is that of a power function, so it makes sense to let $u=2x^3-3.$ Also, if we do so, $du = 6x^2\ dx$, which also appears in the problem.

Making the substitution and integrating: $\int u^4 \ du = \dfrac{1}{5}u^5 + C$

Finally, substitute back to get the answer in terms of $x$: $(2x^3-3)^4 \cdot 6x^2\ dx = \ans{\dfrac{1}{5}(2x^3-3)^5 + C}$

I'll leave it to you to check by differentiating the answer.

Evaluate the following integral: $\displaystyle\int e^{7x} \ dx$.

Solution

What's different about this example? As before, let $u$ be the exponent so that we'll have $e^u$ after substituting: $u = 7x \longrightarrow du = 7 \ dx$

The difference is that now, there's no $7$ in the integral to be replaced with $du$.

What we do have is $dx$, so we can solve for that: $du = 7 \ dx \longrightarrow \dfrac{1}{7} \ du = dx$

After replacing $7x$ with $u$ and $dx$ with $\dfrac{1}{7}\ du$: $\int e^{7x}\ dx = \int \dfrac{1}{7}\ e^u \ du = \dfrac{1}{7}\ e^u + C = \ans{\dfrac{1}{7}\ e^{7x} + C}$

Evaluate $\displaystyle\int (3x^4+7)^3 \cdot 5x^3\ dx$.

Solution

Let $u=3x^4+7$, so $du = 12x^3\ dx$. Notice that $12x^3$ is not exactly the $5x^3$ that we have in the problem, but it's only off by a constant multiple, so we can do the same thing we did in the last example.

The part that is the same is $x^3\ dx$, so solve for that: $du = 12x^3\ dx \longrightarrow \dfrac{1}{12}\ du = x^3\ dx$

Making the substitution: $\int (3x^4+7)^3 \cdot 5x^3\ dx = \int u^3 \cdot 5 \cdot \dfrac{1}{12}\ du$

Finally we can integrate and re-substitute: $\int \dfrac{5}{12}\ u^3\ du = \dfrac{5}{12} \cdot \dfrac{1}{4}\ u^4 + C = \ans{\dfrac{5}{48}(3x^4+7)^4 + C}$

Reciprocal Function

Evaluate $\displaystyle\int \dfrac{4}{4x-3}\ dx$.

Solution

This one looks like the reciprocal function, so let $u=4x-3$. In that case, $du = 4\ dx$, so after substituting, we're left with $\int \dfrac{4}{4x-3}\ dx = \int \dfrac{1}{u}\ du$

Integrating gives the natural log function: $\int \dfrac{1}{u}\ du = \ln u + C = \ans{\ln |4x-3| + C}$

Evaluate $\displaystyle\int \dfrac{1}{2x+7}\ dx$.

Solution

Let $u = 2x+7$, so $du = 2\ dx$. Again, we're off by a constant multiple, so we need to solve for $dx$ as before: $du = 2\ dx \longrightarrow \dfrac{1}{2} \ du = dx$

After substituting: $\int \dfrac{1}{2x+7}\ dx = \int \dfrac{1}{u} \cdot \dfrac{1}{2}\ du = \dfrac{1}{2} \ln |u| + C = \ans{\dfrac{1}{2}\ln |2x+7| + C}$

Evaluate $\displaystyle\int_0^4 \sqrt{2x+1}\ dx$.

Solution 1

Having limits of integration just adds one step at the end of the problem: we can do the indefinite integral as usual, then use the Fundamental Theorem of Calculus at the end.

Let $u=2x+1$, so $du = 2\ dx \longrightarrow \dfrac{1}{2}\ du = dx$. $\begin{aligned} \int \sqrt{2x+1}\ dx &= \int \dfrac{1}{2} \sqrt{u} \ du\ &= \int\dfrac{1}{2} u^{1/2}\ du\ &= \dfrac{1}{2} \cdot \dfrac{2}{3} u^{3/2} + C\ &= \dfrac{1}{3}(2x+1)^{3/2} \end{aligned}$

Now, using the limits of integration: $\dfrac{1}{3}(2x+1)^{3/2} \bigg|_0^4 = \dfrac{1}{3} \cdot 9^{3/2} - \dfrac{1}{3} \cdot 1^{3/2} = \ans{\dfrac{26}{3}}$

Solution 2

There's another option for how to solve this, although it's only a small change: instead of doing the whole substitution process and then evaluate the limits, we can take the limits with us into u-land and evaluate them there.

$u = 2x+1$ When $x=0$, $u = 2(0)+1 = 1$. When $x=4$, $u=9$.

Thus, $\int_0^4 \sqrt{2x+1}\ dx = \int_1^9 \sqrt{u}\ du$

After integrating, we get $\dfrac{1}{3}u^{3/2} \bigg|_1^9 = \dfrac{1}{3} \cdot 9^{3/2} - \dfrac{1}{3} \cdot 1^{3/2} = \ans{\dfrac{26}{3}}$

If you notice, we really did the same work in both cases; all that really changed is where we did each step. Either process gives the same answer, so it's really up to you to choose which you prefer.

A Tricky One

Evaluate $\displaystyle\int x(x-4)^9\ dx$.

Solution

Here again there are two parts, so we could let $u=x$ or let $u=x-4$. Letting $u=x$ wouldn't do anything other than rename the variable; it wouldn't simplify the integral at all.

Thus, let $u=x-4$, so $du=dx$.

Substituting: $\int x(x-4)^9\ dx = \int x \cdot u^9\ du$

The only problem is that there's a lingering $x$, so what can we replace $x$ with so that the integral will only involve $u$? The answer lies in the definition of $u$: if $u=x-4$, $x = u+4$: $\int (u+4)u^9\ du$

This still looks complicated, but there's something we can do now that we couldn't do at the beginning: we can distribute the $u^9$ through the parentheses: $\begin{aligned} \int (u+4)u^9\ du &= \int u^{10} + 4u^9\ du\ &= \dfrac{1}{11}u^{11} + \dfrac{4}{10}u^{10} + C\ &= \ans{\dfrac{1}{11}(x-4)^{11} + \dfrac{4}{10}(x-4)^{10} + C} \end{aligned}$

Another Tricky One

Evaluate $\displaystyle\int \dfrac{x^2}{\sqrt{1-x}}\ dx$.

Solution

Let $u=1-x$, so $x=1-u$ and $-du = dx$.

Substituting: $\begin{aligned} \int \dfrac{x^2}{\sqrt{1-x}}\ dx &= \int \dfrac{(1-u)^2}{\sqrt{u}} \ du\ &= \int \dfrac{1-2u+u^2}{u^{1/2}}\ du\ &= \int u^{-1/2} - 2u^{1/2} + u^{3/2}\ du\ &= 2u^{1/2} - \dfrac{4}{3}u^{3/2} + \dfrac{2}{5}u^{5/2} + C\ &= \ans{2(1-x)^{1/2} - \dfrac{4}{3}(1-x)^{3/2} + \dfrac{2}{5}(1-x)^{5/2}+ C} \end{aligned}$