**Integration by substitution**, also called "u-substitution" (because many people who do calculus use the letter **u** when doing it), is the first thing to try when doing integrals that can't be solved "by eye" as simple antiderivatives.

The idea is to temporarily replace a part of the integral something more simple, perform the integral and re-substitute at the end. You need to work at integration by substitution to build up some intuition about it, but it will happen. Its best just to learn this by example, so here we go:

**Solution**

$$\int \frac{x \, dx}{x^2} = \int \frac{dx}{x} = ln|x| + C$$

Pretty easy. But let's do it again by substitution as a way of figuring the method out.

The substitution is

$$Let \; u = x^2, \; then \; \frac{du}{dx} = 2x$$

Now we rearrange a bit:

Now plugging in **du/2** for **x·dx** and **u** for **x ^{2}**, we can solve the integral

$$ \begin{align} \frac{1}{2} \int \frac{du}{u} &= \frac{1}{2} ln|u| + C \\ \\ &= \frac{1}{2} ln(x^2) + C \; \; \leftarrow \; resubstitute \; u = x^2 \\ \\ &= ln|x| + C \end{align}$$

Solving this particular integral by u-substitution turns out to be a little more difficult than just doing it the regular way, but the technique will be the only thing that will allow you to integrate many kinds of functions.

Now let's apply u-substitution to some more difficult integrands.

The real trick to integration by u-substitution is keeping track of the constants that appear as a result of the substitution. These have to be accounted for, such as the multiplication by ½ in the first example. Watch for that in the examples below.

**Solution****x** in the denominator is one greater than that of the numerator. Because we'll be taking a derivative to do the substitution, the power of what's in the denominator will drop by one to match that of the numerator, and that could work. We'll make this substitution:

Let $u = x^2 + 7,$ then $du = 2x\, dx$

The resulting new integral is

$$\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} ln|u| + C$$

Re-substitution of **x ^{2} + 7** for

$$= \frac{1}{2} ln(x^2 + 7) + C$$

When the result of integration is a **log** (or **ln**) function, we really should write it with absolute-value bars: **ln|x|**, because negative numbers are not included in the domain of **ln(x)**. But if the argument of the log function can't ever be negative, like **ln(x ^{2} + 7)**, then it's OK to leave them off.

**Solution**

$$\int tan(x) \, dx = \int \frac{sin(x)}{cos(x)} \, dx$$

Then we can make this substitution:

Let $u = cos(x),$ then $du = -sin(x) \, dx$

How do we know to make that substitution? Intuition mostly, but if you guess wrong and let **u = sin(x)**, you'll run into a brick wall, but did it really hurt? Probably not. Just try the other way. You didn't hurt the problem; it doesn't have feelings. The resulting simplified integral is

$$- \int \frac{du}{u} = -ln|u| + C$$

Now we can re-substitute **u = cos(x)** to get the result, remembering to re-substitute **cos(x)** for **u**.

$$\int tan(x) \, dx = -ln(cos(x)| + C$$

That's a handy result and one you might want to keep close by. It's good to know the integral of **tan(x)**.

It's always worth a try when working integrals of trig functions to rewrite them using the trig. identities you remember.

**Solution****x ^{2}** and

Let $u = x^3,$ then $du = 3 x^2 \, dx$

The simplified integral and its solution are

$$\frac{1}{3} \int sin(u) \, du = -\frac{1}{3} cos(u) + C$$

Re-substituting for **u** gives us:

$$\int x^2 \, sin(x^3) \, dx = -\frac{1}{3} cos(x^3) + C$$

Whenever the integrand contains powers of **x** of **n** and **n+1**, try substitution.

When considering how to do an integral, look for powers of components that are one apart, like x^3 and x^4. Because taking a derivative drops the power of a function by 1, those are good candidtates.

**Solution****ln(x)** and **1/x**. It's a good clue about how to do the substitution:

Let $u = ln(x),$ then $du = \frac{1}{x} \, dx$

The resulting integral couldn't be simpler:

$$\int u \, du = \frac{1}{2} + C$$

$$\int \frac{ln(x)}{x} \, dx = \frac{1}{2} [ln|x|]^2 + C$$

You can confirm that this, or any of the integrals we've done here are correct by taking the derivative of the result. By the fundamental theorem of calculus, you should get the integrand back.

Consider the definite integral

$$\int_1^3 \frac{x^2 \, dx}{(2 - x^3)^2}$$

**Solution****x ^{3}** and the

Let $u = 2 - x^3,$ then $du = -3x^2 \, dx$

Plugging in we get the simplified integral

$$-\frac{1}{3} \int \frac{du}{u^2} = \frac{1}{3u}$$

Re-substitution of **u = 2 - x ^{3}** gives

Evaluating at the limits is always a little tedious; it gives

$$ \begin{align} &= \frac{1}{3(2 - 27)} - \frac{1}{3(2 - 1)} \\ \\ &= \frac{-1}{75} - \frac{1}{3} = -\frac{13}{38} \end{align}$$

We can stop at the stage where we've solved the simplified integral in terms of **u**, and simply recalculate the limits in terms of **u**

$$-\frac{1}{3} \int \frac{du}{u^2} = \frac{1}{3u}$$

The new limits are calculated in this manner. We know that **u = 2 - x ^{3}**, so when

$$ \begin{align} x = 1 \; &\rightarrow \; u = 2 - 1^3 = 1 \\[5pt] x = 3 \; &\rightarrow \; u = 2 - 3^3 = -25 \end{align}$$

That gives us a new set of limits to evaluate, but this time we leave **x** behind and work with **u**:

$$-\frac{1}{3} \int_1^{-25} \frac{du}{u^2} = \frac{1}{3u} \bigg|_-^{-25}$$

The result is a little simpler to calculate.

$$= \frac{-1}{75} - \frac{1}{3} = -\frac{13}{38}$$

This trick is almost always a simplification, and it's worth learning.

**Solution**

Let $u = (2x - 3),$ then $du = 2 \, dx$

Now we can quickly change the limits of integration, too, and just stick with u from here on:

$$ \begin{align} x = 0 \; &\rightarrow \; u = 2(0) - 3 = -3 \\[5pt] x = 1 \; &\rightarrow \; u = 2(1) - 3 = -1 \end{align}$$

The simplified integral is

$$\frac{1}{2} \int_{-3}^{-1} u^3 \, du = \frac{1}{8} u^4 \bigg|_{-3}^{-1}$$

and the arithmetic of evaluating the limits goes like this.

$$= \frac{(-1)^4}{8} - \frac{(-3)^4}{8} = \frac{1}{8} - \frac{81}{8} = -10$$

See if you can do it by re-substitution of 2x-3 or u and use the original limits. The result should be the same.

When performing u-substitution, it's almost always a good idea to convert the limits of integration to u-based limits.

Solve the following indefinite and definite integrals.

1. | $\int (x - 6)^3 \, dx$ | |

2. | $\int (x + 16)^{1/2}$ | |

3. | $\int \frac{x^3}{(5 - 2x^4)^7} \, dx$ | |

4. | $\int x^2 \sqrt{x^3 - 2} \, dx$ | |

5. | $\int x \, e^{-x^2} \, dx$ |

6. | $\int \frac{sec^2(\sqrt{x})}{\sqrt{x}} \, dx$ | |

7. | $\int_1^7 (x + 3)^3 \, dx$ | |

8. | $\int_0^1 x \, tan(x^2) \, dx$ | |

9. | $\int_{0}^1 \sqrt{3x + 2} \, dx$ | |

10. | $\int_0^2 x \sqrt{5 - x^2} \, dx$ |

Here are three different examples of integration by U-substitution, spanning simple functions, logs and trigonometric functions. The trickiest things about U-substitution are (1) figuring out what to substitute – that's mostly trial-and-error and developing an intuition, and (2) figuring out how to properly substitute for the part that has "**dx**" in it.

Minutes of your life: 4:48

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