# Trigonomic integrals

1. Mar 7, 2012

### ex81

1. The problem statement, all variables and given/known data

do my professor did this in class, and it doesn't make sense to me
∫cos^5(x) sin^4(x) dx
∫cos^4(x) sin^4(x) cos(x) dx
∫(1-sin^2(x))^2 sin^4(x) cos(x) dx
∫(sin^4(x) -2sin^6(x) +sin^8(x))cos(x) dx

so the above I get but when my professor integrated I became lost, this is his solution

1/5 sin^5(x) - 2/7 sin^7(x) +1/9 sin^9(x)

shouldn't those sin be cos? And where did the cos on the outside go?

2. Mar 7, 2012

### tiny-tim

hi ex81!

(try using the X2 button just above the Reply box )
eg (sin9x)' = 9sin8x(sinx)' = 9sin8xcosx

(from the chain rule … (f(g(x))' = f'(g(x))g'(x))

3. Mar 7, 2012

### HallsofIvy

Another way to think about it is that you are making the substitution y= sin(x) so that dy= cos(x) dx. The integral becomes
$$\int (y^4- 2y^6+ y^8)dy$$

integrate that and replace y with sin(x).

4. Mar 11, 2012

### ex81

I think I know where you are going.

Lets try another one :D

∫cos^5(x)/(sin(x))^1/2 dx
so the next step would be to split a cos off of that
∫cos^4(x) cos(x) /(sin(x))^1/2 dx
∫(1-sin^2(x))^2 (cos(x) /(sin(x))^1/2 dx
∫ (1-2sin^2(x) + sin^4(x)) cos(x) /(sin(x))^1/2 dx

This is where I get stuck, and it seems more complicated than it needs to be but I am not sure where to go

5. Mar 11, 2012

### tiny-tim

then eg the ∫ sin4x cosx / sin1/2x dx

= ∫ sin3.5x cosx dx

= … ?

6. Mar 11, 2012

### ex81

Well, that is definitely different than what I was trying to do. Looks way more simple :)

hehe, go figure I wouldn't look at it from a POV of long division

∫(1/(sin(x))^1/2 -2sin^3/2 (x) + sin^7/2 (x) )cos (x)

so that would be 2 sin^1/2 (x) -4/5 sin^5/2 (x) + 2/9 sin^9/2 (x) +C

THANKS!!