Calculus 2 Integral Question (difficult)

[hamburgler]

Homework Statement

$$\int$$cos³xdx

Homework Equations

cos³x = cos²x * cosx

The Attempt at a Solution

Let u= cos²x
du=(x/2) + (cos²x/4x)
v=-sinx
dv=cosuv-$$\int$$vdu

-cos²xsinx - $$\int$$-sinx(x/2 + cos²x/4x)dx

-cos²xsinx -$$\int$$xsinx/2 + (sinxcos²x/4x)

-cos²xsinx - (1/2)xsinx +(sinxcos²x/4x)

-cos²xsinx -(sinx-xcosx/2) + (sinxcos²x/4x)

...aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.

 

[||spoon||]

Hey,

Why don't you try and use the trig identity$$\cos^2x +\sin^2x = 1$$ and then try substituting $$\sin x = u$$

Hope that helps,

-Spoon

 

[hamburgler]

hm ill give it a shot

 

[||spoon||]

No worries, glad to have helped.

- Spoon

 

[HallsofIvy]

Homework Statement

$$\int$$cos³xdx

Homework Equations

cos³x = cos²x * cosx

The Attempt at a Solution

Let u= cos²x
du=(x/2) + (cos²x/4x)

Here's your first error. The derivative of cos² x, using the chain rule, is (2 cos x)(cos x)'= - 2sin x cos x. I have absolutely no idea how you got your du!

v=-sinx
dv=cos


uv-$$\int$$vdu

-cos²xsinx - $$\int$$-sinx(x/2 + cos²x/4x)dx

-cos²xsinx -$$\int$$xsinx/2 + (sinxcos²x/4x)

-cos²xsinx - (1/2)xsinx +(sinxcos²x/4x)

-cos²xsinx -(sinx-xcosx/2) + (sinxcos²x/4x)

...


aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.

You started wrong by trying to use integration by parts when it is not necessary. cos³ x dx= (cos² x) (cos x dx)= (1- sin² x) (cos x dx). Now use the obvious substitution.