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

[tanky322]

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!



Thanks Alot!

 

[Mindscrape]

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.

 

[tanky322]

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?

 

[Noober]

(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?

 

[Office_Shredder]

I would stick with changing cosine to sin. Then you get
\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx

A discerning eye will note a great substitution there

 

[IMDerek]

Perhaps you should try using

\sin (2x) = 2 \sin (x) \cos (x)

edit: sorry, didn't see the new problem

 

[tanky322]

Ok here's my shot at this one at 6:40 AM

so it becomes

\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx

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

 

[Office_Shredder]

Looks good to me