The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Substituting to Simplify the Integral

Date: 08/04/2003 at 12:45:27
From: Trista
Subject: Integration

What is the integral of tan^3 x * secx dx?

Date: 08/04/2003 at 13:28:54
From: Doctor Barrus
Subject: Re: Integration

Hi, Trista.

This is a good question. Finding the indefinite integral of a bunch 
of trigonometric functions is often challenging, and sometimes it 
takes a lot of tries to get something that works.

One strategy that you'll want to adapt is to look for a good 
substitution that will make the integral simpler. I'm going to assume 
that you're familiar with substitutions; let us know if you need more 
information on this.

What I'll do is look at different choices for our new variable u, 
based on what you have already in the integral, and figure out what 
the differential du would be with such a choice...

If           then
u = tan x    du = sec^2 x dx
u = tan^2 x  du = 2(tan x)(sec^2 x) dx
u = tan^3 x  du = 3(tan^2 x)(sec^2 x)dx
u = sec x    du = sec x tan x dx

The first three didn't look promising. All of them required a sec^2 x 
in order to substitute du into the integral. The last one looks a lot 
better, though, because we DO have a sec x tan x dx in the indefinite 

What we'll want to do, then, is to pull this quantity apart from the 
rest of the indefinite integral. In other words, we'll rewrite

(tan^3 x)(sec x)dx


(tan^2 x)(sec x tan x dx)

Remember that it doesn't matter in which order we multiply things. 
Then we know that if we make the substitution u = sec x, we'll be able 
to change the sec x tan x dx into du.

The question now is whether we can write that remaining tan^2 x in 
terms of sec x, so we can replace it by something involving u. What 
relations (identities, etc.) exist between the functions tan x and sec 

I hope you're familiar with one of the Pythagorean identities that 
relates these two:

1 + tan^2 x =sec^2 x

We want to write tan^2 x in terms of sec x, so we'll solve for tan^2 
x in this equation:

tan^2 x = sec^2 x - 1

Or, putting it in terms of u = sec x, we have

tan^2 x = u^2 - 1

We can substitute this into the indefinite integral to arrive at the 
following indefinite integral:

(u^2 - 1) du

and we hope you know how to finish the problem from there.

Does this make sense? Substitution is a good strategy to consider in 
problems like this, particularly in instances where one of the trig 
functions is raised to an odd power, as in this case.

I hope this helps. If you have any more questions, feel free to write 

- Doctor Barrus, The Math Forum 
Associated Topics:
High School Calculus
High School Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.