Calculus 2 Integral Question (difficult)

Click For Summary
SUMMARY

The forum discussion centers on solving the integral \(\int \cos(3x) \, dx\). Participants highlight the importance of using the correct substitution and integration techniques, specifically pointing out that integration by parts is unnecessary for this problem. The recommended approach involves recognizing the identity \(\cos(3x) = \cos(2x) \cdot \cos(x)\) and utilizing the substitution \(u = \sin(x)\) to simplify the integral. Errors in differentiation and substitution methods are also addressed, emphasizing the need for accuracy in calculus techniques.

PREREQUISITES
  • Understanding of trigonometric identities, specifically \(\cos(3x) = \cos(2x) \cdot \cos(x)\)
  • Familiarity with integration techniques, including substitution and integration by parts
  • Knowledge of the chain rule for differentiation
  • Basic skills in manipulating integrals involving trigonometric functions
NEXT STEPS
  • Learn the application of trigonometric identities in integration, focusing on \(\cos(3x)\) and \(\sin(2x)\)
  • Study the method of substitution in integrals, particularly with trigonometric functions
  • Practice integration by parts with various functions to understand when it is necessary
  • Explore common mistakes in calculus, especially in differentiation and integration techniques
USEFUL FOR

Students studying calculus, particularly those tackling integral calculus, as well as educators looking for examples of common pitfalls in integration techniques.

hamburgler
Messages
22
Reaction score
0

Homework Statement


\intcos3xdx

Homework Equations


cos3x = cos2x * cosx

The Attempt at a Solution



Let u= cos2x
du=(x/2) + (cos2x/4x)
v=-sinx
dv=cosuv-\intvdu

-cos2xsinx - \int-sinx(x/2 + cos2x/4x)dx

-cos2xsinx -\intxsinx/2 + (sinxcos2x/4x)

-cos2xsinx - (1/2)xsinx +(sinxcos2x/4x)

-cos2xsinx -(sinx-xcosx/2) + (sinxcos2x/4x)

...aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.
 
Physics news on Phys.org
Hey,

Why don't you try and use the trig identity\cos^2x +\sin^2x = 1 and then try substituting \sin x = u

Hope that helps,

-Spoon
 
hm ill give it a shot
 
No worries, glad to have helped.

- Spoon
 
hamburgler said:

Homework Statement


\intcos3xdx

Homework Equations


cos3x = cos2x * cosx


The Attempt at a Solution



Let u= cos2x
du=(x/2) + (cos2x/4x)
Here's your first error. The derivative of cos2 x, using the chain rule, is (2 cos x)(cos x)'= - 2sin x cos x. I have absolutely no idea how you got your du!

v=-sinx
dv=cos


uv-\intvdu

-cos2xsinx - \int-sinx(x/2 + cos2x/4x)dx

-cos2xsinx -\intxsinx/2 + (sinxcos2x/4x)

-cos2xsinx - (1/2)xsinx +(sinxcos2x/4x)

-cos2xsinx -(sinx-xcosx/2) + (sinxcos2x/4x)

...


aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.

You started wrong by trying to use integration by parts when it is not necessary. cos3 x dx= (cos2 x) (cos x dx)= (1- sin2 x) (cos x dx). Now use the obvious substitution.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K