Why Doesn't the Integral of cos(x)sin(x) Equal (1/2)sin(2x)?

  • Context: Undergrad 
  • Thread starter Thread starter daudaudaudau
  • Start date Start date
Click For Summary
SUMMARY

The integral of cos(x)sin(x) does equal (1/2)sin(2x) when evaluated correctly, but the confusion arises from the integration process. By substituting u = sin(x), the left integral simplifies to (1/2)sin^2(x) + C. The right integral, when evaluated, results in -(1/4)cos(2x) + C', which can be transformed to match the left side by recognizing that cos(2x) can be expressed in terms of sin^2(x). Thus, both integrals are equivalent when constants are adjusted appropriately.

PREREQUISITES
  • Understanding of basic calculus, specifically integration techniques.
  • Familiarity with trigonometric identities, particularly cos(2x) = 1 - 2sin^2(x).
  • Knowledge of substitution methods in integration.
  • Ability to manipulate and simplify algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Study integration techniques involving trigonometric functions.
  • Learn about trigonometric identities and their applications in calculus.
  • Explore substitution methods in depth, particularly in the context of integrals.
  • Practice problems involving the integration of products of sine and cosine functions.
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of integration techniques and trigonometric identities.

daudaudaudau
Messages
297
Reaction score
0
We know that

[tex]\cos{x}\sin{x}=\frac{\sin{2x}}{2}[/tex]

so why isnt

[tex]\int\cos{(x)}\sin{(x)}dx=\int\frac{\sin{(2x)}}{2}dx[/tex]

?
 
Physics news on Phys.org
They are the same. Why do you think they aren't? It's easy to integrate both sides and then show that the integrals are the same.

For example, on the left side, you can let u= sin(x) so that du= cos(x)dx and the integral becomes [itex]\int u du= (1/2)u^2+ C= (1/2)sin^2(x)+ C[/itex]

The integral on the right is simply [itex]-(1/4)cos(2x)+ C'[/itex]. Not obvious that those are the same? cos(2x)= cos2(x)- sin2(x)= (1- sin2(x))- sin2(x)= 1- 2sin2(x) so [itex]-(1/4)cos(2x)+ C'= 1+ (1/2)sin^2(x)+ C'[/itex] which is exactly the same as the first with C= 1+ C'.
 
Ah okay, I see. Thank you.
 

Similar threads

Undergrad Integration of √(1-x^2) with x=sinθ: Check Correctness and Simplification
  • · Replies 12 ·
Replies
12
Views
3K
Undergrad Find ##\int_0^π \sin^ n x dx ##
  • · Replies 5 ·
Replies
5
Views
5K
Undergrad Did Math DF's integral calculator make a glaring mistake?
  • · Replies 8 ·
Replies
8
Views
4K
Undergrad Integration Using Trigonometric Substitution
  • · Replies 8 ·
Replies
8
Views
3K
High School Integration by parts of inverse sine, a solved exercise, some doubts...
  • · Replies 4 ·
Replies
4
Views
3K
High School How Do You Derive the Formula for sin(x-y)?
  • · Replies 5 ·
Replies
5
Views
2K
High School Trigonometric substitution, a case I'd like to share
  • · Replies 3 ·
Replies
3
Views
4K
Undergrad Integration of ##e^{-x^2}## with respect to ##x##
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad Why is the integral of ##\arcsin(\sin(x))## so divisive?
  • · Replies 3 ·
Replies
3
Views
3K