Expert: Paul Klarreich - 2/18/2011

Question
QUESTION: What is the integral of sec^3 (x)? Battling with this one

ANSWER: Questioner:   Peter
Category:  Calculus
Private:  No
 
Question:  What is the integral of sec^3 (x)? Battling with this one


         1
sec^3 (x) = --------
         cos^3 x

  cos x
= --------
 cos^4 x

    cos x
= ----------------
 (1 - sin^2 x)^2

Let u = sin x

       du
= ------------
 (1 - u^2)^2


Now use partial fractions.


---------- FOLLOW-UP ----------

QUESTION: I also have a problem with partial fractions of this form. Would it be possible for you to explain in detail how to do this. I have an important exam coming up. Your help would be greatly aprreciated

Answer
Questioner: Peter
Country: South Africa
Category: Calculus
Private: No
Subject: Trig integral

      du
= ------------
(1 - u^2)^2


Now use partial fractions.


---------- FOLLOW-UP ----------

QUESTION: I also have a problem with partial fractions of this form. Would it be possible for you to explain in detail how to do this. I have an important exam coming up. Your help would be greatly appreciated

Starting  with:
      1  
= ------------
 (1 - u^2)^2

Write:

         1  
= -------------------
 [(1 + u)(1 - u)]^2

         1  
= -------------------
 (1 + u)^2(1 - u)^2

Now the standard method, as I recall it, is:

         1          A         B          C          D
= ------------------- =  -------  + ------- + ---------- + ---------
 (1 + u)^2(1 - u)^2     (1 + u)    (1 - u)   (1 + u)^2    (1 - u)^2

Multiply through by   (1 + u)^2(1 - u)^2  to clear fractions:

1 =  A (1 + u)(1 - u)^2 + B(1 + u)^2(1 - u) + C(1 - u)^2 + D(1 + u)^2

Now substitute four different u's and get four equations for A,B,C,D.  
(I would take  u = 0,1,-1,2, I suppose.)

Over to you, now.  (Why should I have all the fun?)

When you are done, check your answer at:

http://integrals.wolfram.com/index.jsp