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