How do you integrate #int cos^3(x/3)dx#?

Question

Isabella Ackerman · Accepted Answer

#3sin(x/3)-sin^3(x/3)+C# First let #t=x/3#. This implies that #dt=1/3dx#. We then see that: #intcos^3(x/3)dx=3intcos^3(x/3)1/3dx=3intcos^3(t)dt# To do this, split up #cos^3(t)# into #cos^2(t)cos(t)# and then rewrite #cos^2(t)# using the identity #sin^2(theta)+cos^2(theta)=1=>cos^2(theta)=1-sin^2(theta)#. #3intcos^3(t)dt=3intcos^2(t)cos(t)dt=3int(1-sin^2(t))cos(t)dt# Now let #s=sin(t)#, so #ds=cos(t)dt#. Luckily we already have this in the integrand! #3int(1-sin^2(t))cos(t)dt=3int(1-s^2)ds# Integrating term by term using #ints^nds=s^(n+1)/(n+1)# where #n!=-1#. #3int(1-s^2)ds=3(s-s^3/3)=3s-s^3# From #s=sin(t)# and #t=x/3# we see that #s=sin(x/3)#. Also add the constant of integration. #3s-s^3=3sin(x/3)-sin^3(x/3)+C#

Emma Aldridge · Answer

undefined To integrate $ \int \cos^3(\frac{x}{3}) \, dx $, you can use the reduction formula for powers of cosine. The reduction formula for $ \int \cos^n(ax) \, dx $ where $ n $ is a positive integer is:

$$ \int \cos^n(ax) \, dx = \frac{\cos^{n-1}(ax) \sin(ax)}{a} + \frac{n-1}{n} \int \cos^{n-2}(ax) \, dx $$

Applying this formula to $ \int \cos^3(\frac{x}{3}) \, dx $, with $ a = \frac{1}{3} $ and $ n = 3 $, we get:

$$ \int \cos^3(\frac{x}{3}) \, dx = \frac{\cos^2(\frac{x}{3}) \sin(\frac{x}{3})}{\frac{1}{3}} + \frac{2}{3} \int \cos(\frac{x}{3}) \, dx $$

Now, integrate $ \int \cos(\frac{x}{3}) \, dx $ separately to get:

$$ \int \cos(\frac{x}{3}) \, dx = 3 \sin(\frac{x}{3}) + C $$

Finally, substitute this result back into the equation above to find the integral of $ \int \cos^3(\frac{x}{3}) \, dx $.

Adam Adkins · Answer

undefined To integrate ∫cos^3(x/3) dx, you can use the trigonometric identity:

cos^3(x) = (1/4)(3cos(x) + cos(3x))

After applying this identity, you can integrate the resulting expression term by term.