In this section we reverse the chain rule by making a substitution.

### 1 -Substitution

We make a -substitution to fill in the gaps in the following equation, which reverses the chain rule:

If we let be the inside of the composition and we compute : \begin{align*} &\text{Let} \quad \;\;u = g(x) \\ &\text{then} \quad du = g'(x) \ dx \end{align*}

Substituting and into the original integral gives

which we can compute if we know an anti-derivative of : Finally, we can back substitute to get the final answer: Now we can see that

First, note that the function is a composition. The inside function is . Moreover, the derivative of the inside function is which is also present in our integrand. These observations suggest that we should try the substitution . If we let , then using the differential form of the derivative, we have

Now, the original integral which was in terms of the variable and the differential can be rewritten in terms of the new variable and its differential . Substituting for and for , the original integral becomes We now have a basic integral in the variable which can easily be computed: This is essentially our answer except that it is in the wrong variable. We must go back to using : Thus, using -substitution we can conclude that As a final remark, we can check our answer by differentiation: This agrees with our original integrand, so our answer is correct!

Compute: .

Let . Then .

Convert to an integral in the variable : The final answer in terms of is:

and the integral can be written as The last integral can be computed as and by back substituting, we have Thus, using -substitution we can conclude that

Compute .

Let . Then .

Convert to an integral in the variable : The final answer in terms of is:

Thus, using -substitution we can conclude that

Compute .

Let . Then .

If , then and

if , then

Convert to a definite integral in the variable : The final answer is:

The last integral can be computed as and by back substituting, we have Thus, using -substitution we can conclude that

The last integral can be computed as and by back substituting, we have Thus, using -substitution we can conclude that

and the integral can be written as The last integral can be computed as and by back substituting, we have Thus, using -substitution we can conclude that

and the integral can be written as \begin{align*} \int 3x^2\sec (x^3)\tan (x^3) \ dx &= \int \sec (x^3)\tan (x^3) \cdot 3x^2 \ dx\\ &= \int \sec u\tan u \ du. \end{align*}

The last integral can be computed as and by back substituting, we have Thus, using -substitution we can conclude that

and the integral can be written as \begin{align*} \int x^3\cos (x^4) \ dx &= \int \cos (x^4) \cdot x^3\ dx \\ &= \int \cos (x^4)\cdot \tfrac 14 \cdot 4x^3\ dx\\ &= \int \tfrac 14\cos u \ du. \end{align*}

Step 1) Transform the integral: \begin{align*} \int \frac{1}{x\ln ^2 x}& \ dx \, = \, \int \frac{1}{\ln ^2 x} \cdot \frac{1}{x} \, dx\, = \, \int \frac{1}{u^2} \ du\\ \text{let} \;\; u = & \ \ln (x)& \\ \text{then} \;\; du = & \ \frac{1}{x} \ dx& \end{align*}

Step 2) Anti-differentiate: Step 3) Back substitute: Hence,

Let . Then we have, and the integral can be written as The last integral can be computed as and by back-substituting, we have Thus, using -substitution we can conclude that

The last integral can be computed as and by back-substituting, we have Thus, using -substitution we can conclude that