High School Calculus/Integration by Substitution

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Integration by Substitution[edit]

There is a theorem that will help you with substitution for integration. It is called Change of Variables for Definite Integrals.

what the theorem looks like is this

\int_{a}^{b} f(x)\operatorname {d}x = \int_{\alpha}^{\beta} f(g(u))g\prime (u)\operatorname {d}u


In order to get \alpha you must plug a into the function g and to get \beta you must plug b into the function g.

The tricky part is trying to identify what you want to make your u to be. Some times substitution will not be enough and you will have to use the rules for integration by parts. That will be covered in a different section

Ex. 1

\int_{0}^{2} x(x^2+1)^2 \operatorname {d}x

Instead of making this a big polynomial we will just use the substitution method.

Step 1

Identify your u

Let  u = x^2+1

Step 2


Identify \operatorname {d}u


\operatorname {d}u = 2x\operatorname {d}x


Step 3

Now we plug in our limits of integration to our u to find our new limits of integration

When  x = 0, u =0^2 + 1 = 1

and when x = 2, u = 2^2 + 1 = 5

Now our integration problem looks something like this

\frac {1}{2} \int_{0}^{5} (x^2 + 1)^2 (2x)\operatorname {d}x

Step 4

write your new integration problem


When we plug in our u it looks like

\frac {1}{2} \int_{0}^{5} (u)^2 \operatorname {d}u


Step 5

Evaluate the Integral

\frac {1}{2} \left[\frac {1}{3} u^3 \right]_{0}^{5}


\frac {1}{2} \left[\left(\frac {1}{3} * 5^3 \right) - \left(\frac {1}{3} * 0^3 \right)\right]



\frac {1}{2} \left[\frac {1}{3} * 125 \right]



\frac {1}{2} \left[\frac {125}{3}\right]



\frac {125}{6}



As you can see this all simplified fairly nice. Using substitution will be hard, for most people, at first. Once you get the hang of doing this it should come to you faster and faster each time.

I'll give you some other problems to work on as well.

Ex. 2

\int_{0}^{\frac {\pi}{2}} \sin (x) \cos (x) \operatorname {d}x

Ex. 3

\int_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x

Solutions[edit]

Ex. 2

Let u = \sin (x)

Then \operatorname {d}u = \cos (x) \operatorname {d}x

When x = 0
\sin (0) = 0
and when x = \frac {\pi}{2}
\sin \left(\frac {\pi}{2}\right) = 1


Therefore,

\int_{0}^{1} u \operatorname {d}u


\left[\frac {1}{2} u^2 \right]_{0}^{1}

\left[\frac {1}{2} \sin ^2 (1) \right] - \left[\frac {1}{2} \sin ^2 (0) \right]

= \frac {1}{2} \sin ^2 (1)

Ex. 3

Let u = x^2 + 4

Then \operatorname {d}u = 2x

plug in our limits to get new limits

When x = -1
(-1)^2 + 4 = 5
and when x = 2
2^2 + 4 = 8

Our new integration problem is

\int_{5}^{8} (u)^\frac {1}{2} \operatorname {d}u


Giving us

\left[\frac {2}{3} u^\frac {3}{2} \right]_{5}^{8}

= \left[\frac {2}{3} * (x^2 + 4) \right]_{5}^{8}

= \left[\frac {2}{3} * (8^2 +4)\right] - \left[\frac {2}{3} * (5^2 +4) \right]

= \left[\frac {136}{3}\right] - \left[\frac {58}{3} \right]

= 26