How do you integrate ((sin x)^5) * ((cos x)^3) in the limits pi/2 to 3pi/4?

Question

This is not a homework question. My colleague at work has started tutoring someone in math and asked me this question to see if I could help, but I have no clue.

Answer 1

The correct integral is:

(-3/128)cos(2x)+(1/256)cos(4x)

+(1/384)cos(6x) - (1/1024)cos(8x)

This link explains how to deal with products of powers of sines and cosines:

http://www.sosmath.com/calculus/integration/powerproduct/powerproduct.html

Answer 2

This one is rather a difficult one to go thru the whole procedure in details, but this is the answer for an equation like this"

Integral of ((cosax)^m)*((sinax)^n)dx =

((cosax)^m-1)*((axsinax)^n+1)/(m+n)a plus (m-1)/(m+n)Integral ((cosax)^m-2)((sinax)^n)dx.

Substitute the numbers knowing that a =1, and m=5, and n= 3
and then you have to take the limits from pi/2 to 3pi/4.

Good luck

Answer 3

Use the trig identity: (cos x)^2 + (sin x)^2 = 1
(cos x)^2 = 1 - (sin x)^2

factor out (cos x)^2 from your original equation and replace it with 1- (sin x)^2. You'll be left with cos x times some equation with only sines in it. So, You can then do a u substitution where u = sinx and du = cos x dx . It shoudl be pretty straightforward from there.

Answer 4

First u get a big glass of milk and add some chocolate. Then you take a Vicodin and close the book.

Answer 5

I believe there is a formula which integrates powers of sines and cosines

Use this and perform the substitutions:

http://integrals.wolfram.com/index.jsp

Answer 6

That is so easy, my child in the 8th grade figured it out. The answer is 8!!

Answer 7

please see http://archives.math.utk.edu/visual.calculus/4/recursion.1/index.html ; it has a similar problem... applying limits is easy thereafter...

in fact, there are generalised formulae for such trigonometric definite integrals...