How do I solve this odd power of sine integral problem?

Just remember to add a "+ C" to the end of the integrals :)In summary, the given integral can be solved by using the identity 1-sin^2(x) = cos^2(x). By making the substitution u = sin(x), the integral can be written as (sin(x)^3)/3 - (sin(x)^5)/5 + C.
  • #1
tanky322
43
0

Homework Statement



Integral((sin(x))^2((cos(x))^2) dx


Homework Equations





The Attempt at a Solution



Seperate Cos (x)^3

sin(x)^2 (cos(x))(cos(x))^3

Then:apply identities

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


And now I am lost!:eek:



Thanks Alot!
 
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  • #2
Is that supposed to be (cosx)^4 or something (unless you divide by cos as well, you cannot have a cos^3 out of cos^2)?

The way I would do this is to use the identity that sin^2 x is 1 - cos^2 x, and then you can integrate cos^n x with a reduction formula or double angle identities.

Or you could start off with double angle identities and see where it leads you.
 
  • #3
Yea Sorry I screwed that up the original integral is
Integral((sin(x))^2((cos(x))^3) dx


Then
Seperate Cos (x)^3

sin(x)^2 (cos(x))(cos(x))^2

Then:apply identities

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

So now If I change the sin(x)^2 to 1-cos(x)^2 I won't have dx just as if I change it back?
 
  • #4
(sinx)^2 = 1-(cosx)^2
so...

Int((sin(x))^2((cos(x))^3) dx
=Int[(cosx)^3(1-(cosx)^2)]dx
=Int[(cosx)^3 - (cosx)^5]dx
= -(cosx)^4/4sinx + (cosx)^6/6sinx

Right?
 
Last edited:
  • #5
I would stick with changing cosine to sin. Then you get
[tex]\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx[/tex]

A discerning eye will note a great substitution there
 
  • #6
Perhaps you should try using

[itex]\sin (2x) = 2 \sin (x) \cos (x)[/itex]

edit: sorry, didn't see the new problem
 
  • #7
Ok here's my shot at this one at 6:40 AM :bugeye:

so it becomes

[tex]\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx[/tex]

by using 1-sin(x)^2

So I make u sin(x)
du cos(x)

Therefore I get (sin(x)^3)/3-(sin(x)^5)/5

That sound about right?

Thanks,
Andy
 
  • #8
Looks good to me
 

1. What is an odd power of sine integral?

An odd power of sine integral is a type of mathematical function that involves the integral of an odd power of the sine function. This means that the power of the sine function is an odd number, such as 3, 5, 7, etc.

2. How is an odd power of sine integral calculated?

An odd power of sine integral is calculated using integration techniques, such as substitution, integration by parts, or trigonometric identities. The specific method used will depend on the form of the integral.

3. What is the purpose of studying odd power of sine integrals?

Studying odd power of sine integrals can be useful in various areas of mathematics, such as solving differential equations, evaluating definite integrals, and finding the area under a curve. It also helps in understanding the behavior of the sine function.

4. Are there any special properties of odd power of sine integrals?

Yes, there are several special properties of odd power of sine integrals. For example, the integral of an odd power of sine function over one full period is always equal to zero. Additionally, the integral of an odd power of sine function over a half period is equal to half the integral over the full period.

5. Can odd power of sine integrals be solved analytically?

Yes, some odd power of sine integrals can be solved analytically using standard integration techniques. However, there are some integrals that cannot be solved analytically and require numerical methods or approximations.

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