The Math Forum

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

Indefinite Integrals

Date: 07/04/2002 at 14:51:38
From: Anirban Bhattacharyya
Subject: Indefinite Integral

1. Integral dx/(sin x)^5 + (cos x)^5.
2. Integral dx/(sin x)^3 + (cos x)^5.

Date: 07/10/2002 at 10:52:02
From: Doctor Nitrogen
Subject: Re: Indefinite Integral

Hello, Anirban:

The two integrals you submitted:

  1. Integral dx/((sinx)^5 + (cosx^5)),

  2. Integral dx/((sinx^3) + (cosx)^5),

can be solved by using various trigonometric identities and by then 
employing certain substitutions. Let's start with (1). 

First, notice that 

  (sinx)^5 + (cosx)^5 = (cosx)^5(1 + (tanx)^5),

using the identity 

  (sinx)^5/(cosx)^5 = (tanx)^5. 

Here all you have done is factored out (cosx)^5, which leaves     
  1 + (sinx)^5/(cosx)^5 

in the parentheses. So now the integral becomes

  Integral dx/(cosx)^5(1 + (tanx)^5).

Using the fact that 

  1/(cosx)^5 = (secx)^5, 

change the integral to:

  Integral (secx)^5dx/(1 + (tanx)^5).

Here is where you can try a substitution. Let

     U = tanx

    dU = (secx)^2dx

  cosx = 1/sqrt(1 + U^2)

  secx = sqrt(1 + U^2).

This changes the integral to:

    Integral (secx)^5dx/(1 + (tanx)^5)

  = Integral (sqrt(1 + U^2))^3dU/(1 + U^5)

  (secx)^5dx = (secx)^3*(secx)^2dx


  dU = (secx)^2dx.

Now notice that 

  (sqrt(1 + U^2))^3 = (1 + U^2)^(3/2.)

Unfortunately, you are not done yet. You must use two more 
substitutions. For the first substitution try

   V = 1 + U^2

   U = sqrt(V - 1)

  dV = 2UdU

  dU = 1/2(dV/U) 

     = 1/2(dV/sqrt(V - 1))

This changes the integral to:

    (1/2) * Integral (V^3/2)dV/(sqrt(V - 1))^6 

  = (1/2) * Integral(V^3/2)dV/(V - 1)^3.

The second substitution is:

  V^1/2 = W

      V = W^2

     dV = 2WdW, 

so that you finally come up with the integral:

  Integral W^4dW/(W + 1)^3(W - 1)^3.

From this point on, you must use partial fractions to get your answer. 

You can solve (2) by a similar method, only it leads to a 
different numerator:


  (sinx)^3 + (cosx)^5, 

factor out cosx^5 in the denominator. For the denominator you will get 

    (cosx)^5(1 + (sinx)^3/(cosx)^5)

  = (cosx)^5(1 + (tanx)^3*(secx)^2).

Use the identity 

  (tanx)^2 = (secx)^2 - 1 

to change the integral to: 

   Integral (secx^5)dx/(1 + (tanx)^3*(secx)^2).

Now use the substitution 

     U = tanx

    dU = secx^2dx

  secx = sqrt(1 + U^2), 

and further substitutions similar to those in the first solution to 
get your answer.

You will again have to use partial fractions after you make the last 
substitutions. And don't forget to "backtrack" through all the 
substitutions you made when you get the integral value in other 
'dummy' variables, so you will wind up again with all expressions 
in x. 

I hope this was helpful. If you have any more questions, please 
contact Dr. Math again.

- Doctor Nitrogen, The Math Forum 
Associated Topics:
College Calculus
High School Calculus

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.