Lecture in Integration by Substitution

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(Last Updated On: December 21, 2017)
Integration by Substitution
If a given function seems to be complex and it is not recognizable as a standard integral. There are techniques that can help us in particular situations. In this case we will be using one such technique called 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

Lecture in Integration by Substitution

Consider this example:

I want you to visualize the integrand as ∫e f (x) ⋅ f’ (x) dx . Let u = f (x) , then the differential du equals f ‘(x)dx , and you have the following.
Let
then,

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?

At this point, you might be asking how to identify the correct substitution. The answer is it depends on the integral. The success in integration rely on and your ability to spot what part of the integrand should be called 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?

After performing the substitution for 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 dx in
    terms of du.
  • 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

For = , n Lecture in Integration by Substitution -1

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

Lecture in Integration by Substitution       and       Lecture in Integration by Substitution

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

Lecture in Integration by Substitution

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

Lecture in Integration by Substitution  for a > 0

Lecture in Integration by Substitution, for a > 0

Lecture in Integration by Substitution, 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

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