# integrate cos^3(x)

• Jul 6th 2009, 12:22 AM
slaypullingcat
integrate cos^3(x)
Hi, how do I integrate cos^3(x)?

I split it into cos(x)[1/2(1+cos(2x)] and then removed the constant of a 1/2 from the integration. Then I mutilpied through by cos(x) and got cos(x)+cos^2(2x), which didn't really help.

Then I split cos^3(x) into cos(x).cos^2(x) tried substituting cos^2(x) with 1-sin^2(x) and that didn't help either.

What am I missing?
Thank you.
• Jul 6th 2009, 12:29 AM
Chris L T521
Quote:

Originally Posted by slaypullingcat
Hi, how do I integrate cos^3(x)?

I split it into cos(x)[1/2(1+cos(2x)] and then removed the constant of a 1/2 from the integration. Then I mutilpied through by cos(x) and got cos(x)+cos^2(2x), which didn't really help.

Then I split cos^3(x) into cos(x).cos^2(x) tried substituting cos^2(x) with 1-sin^2(x) and that didn't help either.

What am I missing?
Thank you.

Note that $\int\cos^3x\,dx=\int\cos^2x\cos x\,dx=\int\left(1-\sin^2x\right)\cos x\,dx$ $=\int\cos x\,dx-\int\sin^2x\cos x\,dx$

In the second integral, make the substitution $u=\sin x$.

Can you continue?
• Jul 6th 2009, 12:34 AM
slaypullingcat
Quote:

Originally Posted by Chris L T521
Note that $\int\cos^3x\,dx=\int\cos^2x\cos x\,dx=\int\left(1-\sin^2x\right)\cos x\,dx$ $=\int\cos x\,dx-\int\sin^2x\cos x\,dx$

In the second integral, make the substitution $u=\sin x$.

Can you continue?

I will certainly try. Thank you for your post.
• Jul 6th 2009, 12:35 AM
Prove It
Quote:

Originally Posted by slaypullingcat
Hi, how do I integrate cos^3(x)?

I split it into cos(x)[1/2(1+cos(2x)] and then removed the constant of a 1/2 from the integration. Then I mutilpied through by cos(x) and got cos(x)+cos^2(2x), which didn't really help.

Then I split cos^3(x) into cos(x).cos^2(x) tried substituting cos^2(x) with 1-sin^2(x) and that didn't help either.

What am I missing?
Thank you.

Use the identity

$\cos^3{x} = \frac{3}{4}\cos{x} + \frac{1}{4}\cos{(3x)}$.

So

$\int{\cos^3{x}\,dx} = \int{\frac{3}{4}\cos{x} + \frac{1}{4}\cos{(3x)}\,dx}$

$= \frac{3}{4}\sin{x} + \frac{1}{12}\sin{(3x)} + C$.
• Jul 6th 2009, 12:39 AM
slaypullingcat
Quote:

Originally Posted by Chris L T521
Note that $\int\cos^3x\,dx=\int\cos^2x\cos x\,dx=\int\left(1-\sin^2x\right)\cos x\,dx$ $=\int\cos x\,dx-\int\sin^2x\cos x\,dx$

In the second integral, make the substitution $u=\sin x$.

Can you continue?

Out of curiosity, is there a method of solving this problem without using substitution. In the cousre I am doing substitution has already been covered so I am fine with that, but I am trying to get the big picture of integration and get a feel for as many techniques I can. Cheers
• Jul 6th 2009, 01:12 AM
Prove It
Quote:

Originally Posted by slaypullingcat
Out of curiosity, is there a method of solving this problem without using substitution. In the cousre I am doing substitution has already been covered so I am fine with that, but I am trying to get the big picture of integration and get a feel for as many techniques I can. Cheers

Yes, use the identity I've given you. There's no $u$ substitution there...