# integral of cos^3(x)

• Aug 20th 2009, 02:23 AM
Mr.Ree
integral of cos^3(x)
Hi, I need to integrate cos^3(x).

So far I have:∫cos^3(x) = ∫cos(x)(1-sin^2(x)) = ∫cos(x) -∫cos(x)sin^2(x)= sin(x) - (1/3) sin^3(x) = 1/3 (3sin(x) - sin^3(x))

How do I get from this to: 1/12 (9sin(x) + sin(3x))

Is there a special trig identity?
• Aug 20th 2009, 03:32 AM
mr fantastic
Quote:

Originally Posted by Mr.Ree
Hi, I need to integrate cos^3(x).

So far I have:∫cos^3(x) = ∫cos(x)(1-sin^2(x)) = ∫cos(x) -∫cos(x)sin^2(x)= sin(x) - (1/3) sin^3(x) = 1/3 (3sin(x) - sin^3(x))

How do I get from this to: 1/12 (9sin(x) + sin(3x))

Is there a special trig identity?

$\displaystyle \sin (3x) = \sin(x + 2x) = \sin x \cos (2x) + \cos x \sin (2x) = \sin x (1 - 2 \sin^2 x) + 2 \sin x \cos^2 x$

$\displaystyle = \sin x - 2 \sin^3 x + 2 \sin x (1 - \sin^2 x) = \sin x - 2 \sin^3 x + 2 \sin x - 2 \sin^3 x$

$\displaystyle = 3 \sin x - 4 \sin^3 x$.

Therefore $\displaystyle \sin^3 x = \frac{3}{4} \sin x - \frac{1}{4} \sin (3x)$.