# Integration Example: Sines an Cosines

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx = ∫ du*(1-u2)2*u4

Or, rewrite it: ∫(1-u2)2*u4du

Simplify and clean up:

∫(1-2u2+u4)*u4du

∫u4-2u6+u8du

Integrate to get:

(1/5)u5-(2/7)u7+(1/9)u9+C

And, substitute back in u = sin(x):

(1/5)sin5(x)-(2/7)sin7(x)+(1/9)sin9(x)+C

This is only 1 way to solve this problem.  I could’ve picked a different u-substitution and I could’ve likely picked different trig identities to help me along the way.

# Integration Examples

Q:

Integrate:  14*(12 – 4x)(12x – 2x2)6 dx

Since I see that (12 – 4x) is the derivative of (12x – 2x2), my instinct tells me to use u-substitution.

So, let

u = 12x – 2x2

then, compute du:

du = 12 – 4x dx

Now, we can directly substitute in u and du into our problem:

Integrate:  14*(12 – 4x)(12x – 2x2)6 dx = 14*u6 du

[the u replaced (12x – 2x2) and the du replaced the (12 – 4x) dx]

So, we are integrating:  14*u6 du

Using the power rule:

int[ 14*u6 du] = 14/7 * u7 + C = 2u7 + C

And since u = (12x – 2x2), the final answer is:

(12x – 2x2)7 + C