Integrals - the Substitution Rule with sin^n(x)

  • #1
sugarxsweet
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0

Homework Statement


Given that n is a positive integer, prove ∫sin^n(x)dx=∫cos^n(x)dx from 0 -> pi/2


Homework Equations


Perhaps sin^2(x)+cos^2(x)=1? Not sure.


The Attempt at a Solution


I honestly don't even know where to start. Should I set u=sin(x) or cos(x)? Doesn't seem to get the right answer either way...
 

Answers and Replies

  • #2
Simon Bridge
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You need to show that the area under the graph from 0 to 90deg is the same for sin and cosine to any power.

Your first step is to understand the problem - try sketching the graphs for a few powers and shading the area in question to see what you are up against.

There is a rule for integrating powers of sin and cos ... you can derive it from integration by parts. Knowing the rule, you can probably just do it algebraically.

Or you can try a substitution in one like [itex]x=\frac{\pi}{2}-u[/itex] and exploit the parity of the functions.
 
  • #3
sugarxsweet
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Sorry, to clarify, the hint says:
Use a trigonometric identity and substitution. Do not solve the definite integrals

Given this information, how would you recommend solving?
 
  • #4
SammyS
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Homework Statement


Given that n is a positive integer, prove ∫sin^n(x)dx=∫cos^n(x)dx from 0 -> pi/2

Homework Equations


Perhaps sin^2(x)+cos^2(x)=1? Not sure.

The Attempt at a Solution


I honestly don't even know where to start. Should I set u=sin(x) or cos(x)? Doesn't seem to get the right answer either way...

The co-function identity states that [itex]\displaystyle \cos(x) = \sin\left(\frac{\pi}{2}-x\right)\ .[/itex]
 
  • #5
sugarxsweet
12
0
Thanks! I got so far:

∫cos^n(x)dx from 0 to pi/2
=∫sin^n(pi/2-x)dx from 0 to pi/2
=∫sin^n(-x)dx from -pi/2 to 0
=∫sin^n(x)dx from 0 to pi/2

Does this look right to you? Thanks for the hint!
 
  • #6
Simon Bridge
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Yeah, that's the idea - it was the second hint post #2 :)

You should comment each step in your actual answer, to explain what you are doing.
 
  • #7
SammyS
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Thanks! I got so far:

∫cos^n(x)dx from 0 to pi/2
=∫sin^n(pi/2-x)dx from 0 to pi/2
=∫sin^n(-x)dx from -pi/2 to 0
=∫sin^n(x)dx from 0 to pi/2

Does this look right to you? Thanks for the hint!
Use the substitution, [itex]u=\frac{\pi}{2}-x[/itex] to show that [itex]\displaystyle \int_{x=0}^{x=\pi/2}\sin^n(\frac{\pi}{2}-x)\,dx = \int_{u=\pi/2}^{u=0}-\sin(u)\,du\ .[/itex]
 
  • #8
sugarxsweet
12
0
Thanks!
 

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