How do you find the antiderivative of # cosx^6#?

Question

Aurora Alvarado · Accepted Answer

#int\ cos^6(x)\ dx=5/16x+1/4sin(2x)+3/64sin(4x)-1/48sin^3(2x)+C#

I don't think that #cos(x^6)# has an antiderivative expressible with elementary functions, so I assume you want to find #int\ cos^6(x)\ dx#.

First, recall that #cos(2theta)=2cos^2(theta)-1#, or #cos^2(theta)=(1+cos(2theta))/2#.

So, we have #\ \ \ \ \ \ int\ cos^6(x)\ dx# #=int\ (cos^2(x))^3\ dx# #=int\ ((1+cos(2x))/2)^3\ dx# #=1/8int\ (1+cos(2x))^3\ dx# #=1/8int\ (1+3cos(2x)+3cos^2(2x)+cos^3(2x))\ dx# #=1/8(int\ dx+3int\ cos(2x)\ dx+3int\ cos^2(2x)\ dx+int\ cos^3(2x))\ dx)#

Let's examine each integral separately.

The first integral is obvious: #\ \ \ \ \ \ int\ dx=x+C#

The second integral requires the reverse chain rule: #\ \ \ \ \ \ 3int\ cos(2x)\ dx# #=3int\ cos(2x)dx/(d(2x))\ d(2x)# #=3/2int\ cos(2x)\ d(2x)# #=3/2sin(2x)+C#

The third integral requires using #cos^2(theta)=(1+cos(2theta))/2# again: #\ \ \ \ \ \ 3int\ cos^2(2x)\ dx# #=3int\ (1+cos(4x))/2\ dx# #=3/2int\ (1+cos(4x))\ dx# #=3/2x+3/2int\ cos(4x)\ dx# #=3/2x+3/2int\ cos(4x)dx/(d(4x))\ d(4x)# #=3/2x+3/8int\ cos(4x)\ d(4x)# #=3/2x+3/8sin(4x)+C#

The fourth integral requires the knowledge that #sin^2(theta)+cos^2(theta)=1# and the reverse chain rule again: #\ \ \ \ \ \ int\ cos^3(2x)\ dx# #=int\ cos(2x)(1-sin^2(2x))\ dx# #=int\ cos(2x)(1-sin^2(2x))dx/(d(sin(2x)))\ d(sin(2x))# #=1/2int\ (1-sin^2(2x))\ d(sin(2x))# #=1/2sin(2x)-1/6sin^3(2x)+C#

Put all of them together to get #5/16x+1/4sin(2x)+3/64sin(4x)-1/48sin^3(2x)+C#

Ezekiel Alexander · Answer

undefined To find the antiderivative of $ \cos^6(x) $, you can use the reduction formula for powers of cosine, which is derived from integration by parts.

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

Using this reduction formula, for $ \int \cos^6(x) \, dx $, we can start by letting $ n = 6 $:

$$ \int \cos^6(x) \, dx = \frac{\cos^5(x) \sin(x)}{6} + \frac{5}{6} \int \cos^4(x) \, dx $$

Now, for the remaining integral $ \int \cos^4(x) \, dx $, we can again apply the reduction formula:

$$ \int \cos^4(x) \, dx = \frac{\cos^3(x) \sin(x)}{4} + \frac{3}{4} \int \cos^2(x) \, dx $$

Since $ \cos^2(x) = \frac{1 + \cos(2x)}{2} $, we substitute this back into the equation:

$$ \int \cos^4(x) \, dx = \frac{\cos^3(x) \sin(x)}{4} + \frac{3}{8}x + \frac{3}{8}\sin(2x) + C $$

Now, substituting this back into the original integral, we get:

$$ \int \cos^6(x) \, dx = \frac{\cos^5(x) \sin(x)}{6} + \frac{5}{6} \left( \frac{\cos^3(x) \sin(x)}{4} + \frac{3}{8}x + \frac{3}{8}\sin(2x) \right) + C $$

This can be further simplified if needed.