*integration by substitution*. We may try changing the variable

*x*by a suitable substitution in terms of another variable,

*u*. The method of u-substitution is a method for algebraically simplifying the form of a function so that its antiderivative can be easily recognized. This method is intimately related to the chain rule for differentiation.

### The Substitution Rule

Consider this example:

*(x)*⋅ f’ (

*x*)

*dx*. Let

*u*= f (

*x*) , then the differential

*du*equals f ‘(

*x*)

*dx*, and you have the following.

Now, you substitute u and the differential *du* into the integrand and integrate.

Lastly, re-substitute for *u* so the integral is in terms of the original variable, *x*.

### How can you identify the correct substitution?

*u*for you to have a correct

*du*, so that a known formula can be applied. The key is the only way to really learn how to do substitutions well is to just work lots of problems.

### How do you know if you got the correct substitution?

*x*in the integral (including the

*x*in the

*dx*), the only letters left should be u’s (including a du). In some cases, where after the correct normal substitution will actually leave some

*x*’s. You are going to need a little more work to substitute every

*x*’s in terms of

*u*.

### Simple steps to determine the correct substitution

- Try to find good
*u*in the integrand. It might be in parentheses or square root. - Find the derivative of u, du/dx, and then by `cross-multiplying’ express d
*x*in

terms of d*u*. - Substitute all
*x*variable with*u*. You may need to use some algebra to express the remaining*x*in terms of*u*, but commonly everything is in terms of*u*. - Integrate in terms of
*u*. After getting the integrals, replace*u*by the expression in*x*it stand for in the first step.

### Integration by substitution with algebraic functions

1.

Let u = x2 + 3x

Finding the differential we get, du = (2x + 3) dx

Substitute into the original problem, replacing all forms of *x*, getting

= =

Replacing back u by x2 + 3x

=

### Integration by substitution involving exponential functions

1. I already give example for the first formula above. Here is for the second,

Let u = 3x

then, du = 3 dx,

⅓ du = dx

= ⅓ =

### Integration by substitution involving trigonometric functions

1. Integrate

Let u = 3x

then du = 3 dx,

⅓ du = dx

Substitute into the original problem, replacing all forms of x, getting

= 2 = 2sin(u) + C = 2sin3x + C

### Algebraic substitution involving integral giving inverse trigonometric functions

, for a > 0

, for u > a > 0

1.Integrate

Let u = tan x

then du = sec2x dx,

Substitute into the original problem, replacing all forms of x, getting

= =

### Problems set exercises for Integration by Substitution

Follow this link to challenge yourself to work on different kinds of problems. Good luck. Integration by Substitution – Set 1 Problems

*credit: D. A. Kouba (UC Davis), Renato E. Apa-ap, et al.**©2013 www.PinoyBIX.com*