What is the issue with u-Substitution in Integrating tan x sec^4 x dx?

  • Thread starter Thread starter andrewjohnsc
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on the issue of u-Substitution in the integral of tan(x) sec^4(x) dx. Two methods of integration were presented, with the first method yielding the correct result of sec^4(x)/4 + C, while the second method produced a different expression, tan^2(x)/2 + tan^4(x)/4 + C'. The discrepancy arises from the constants of integration, which are not equivalent. Both answers are valid, but they differ by a constant, highlighting the importance of recognizing that C and C' represent different constants.

PREREQUISITES
  • Understanding of u-Substitution in calculus
  • Familiarity with trigonometric identities, specifically sec^2(x) = 1 + tan^2(x)
  • Knowledge of integration techniques for polynomial expressions
  • Ability to manipulate constants of integration in indefinite integrals
NEXT STEPS
  • Review the method of u-Substitution in calculus
  • Study trigonometric identities and their applications in integration
  • Practice integrating functions involving secant and tangent
  • Explore the concept of constants of integration and their implications in indefinite integrals
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking to clarify common misconceptions about u-Substitution and trigonometric integrals.

andrewjohnsc
Messages
7
Reaction score
0

Homework Statement


\inttan x sec^{}4 x dx

Homework Equations


U-Sub

The Attempt at a Solution


I did this two different ways and go two different answers.

The correct way is
\intsec^3(x)*sec(x)*tan(x) dx
letting u= sec(x) and du= sec(x)*tan(x)
= \int u^3 du = sec^4(x)/4 + C
Which is good,

but I can't figure out what's wrong with this way:
\inttan(x)*sec^4(x) dx
= \int tan(x)*(1+tan^2(x))(sec^2(x)) dx
from the identity sec^2(x) = 1 + tan^2(x)
then letting u= tan(x) and du= sec^2(x) dx
gives \int u*(1 + u^2) du = \int u + u^3 du = u^2/2 + u^4/4 + C
= tan^2(x)/2 + tan^4(x)/4 + C

Which is a different answer from the correct one. What's wrong with the second way? Does it have to do with the u-sub... Thanks,
Andrew
 
Last edited:
Physics news on Phys.org
Both answers are correct, but the constant C in your second answer isn't the same as the constant C in your first answer.

<br /> \tan^2(x)/2 + \tan^4(x)/4+C&#039;=\frac{1}{4}(\sec^4(x)-1)+C&#039;=\sec^4(x)/4-1/4+C&#039;=\sec^4(x)+A<br />

I leave the intermediate steps to you.
 
sec4x=(1+tan2)2

sec^4x= 1 + 2tan^2x+tan^4x

Now divide by 1/4 and you'll see that they are the exact same thing.

AND 1/4+C = another constant!
 
Thanks a lot!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K