NOTES ON HOW TO INTEGRATE EVEN POWERS OF SINES AND COSINES

NOTES ON HOW TO INTEGRATE EVEN POWERS OF SINES AND COSINES

1. Even powers of cos x

We will begin with an algorithm for writing down an identity for an even power of cos(x). This turns something like

cos8(x) dx

into an equivalent expression ready to integrate. After giving the algorithm, we will

elaborate on where it comes from, and further tips for using the ideas to help set up

integrals. We will take the example of cos6(x). For other even powers, you can just substitute

the power in place of the number 6 in the description below.

Start by writing out Pascal's triangle down to the row that starts with 1 6: 1

11 121 1331 14641 1 5 10 10 5 1 1 6 15 20 15 6 1

Now write down

(20, 30, 12, 2).

This comes from starting in the center on the bottom line of Pascal's triangle above, and then working outward. The middle entry, 20, goes first. Then you add the two 15's to the left and right of 20. Then you add the two 6's, then the two 1's.

Now the identity is:

(2 cos x)6 = 20 cos(0x) + 30 cos(2x) + 12 cos(4x) + 2 cos(6x).

(Don't forget the 2 before the first cosine!)

Since cos(0x) = cos(0) = 1 and (2 cos x)6 = 26 cos6 x = 64 cos6 x, the above

translates to

cos6 x =

1 (20 + 30 cos(2x) + 12 cos(4x) + 2 cos(6x))

64

1 = (10 + 15 cos(2x) + 6 cos(4x) + cos(6x)).

32

1

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NOTES ON HOW TO INTEGRATE EVEN POWERS OF SINES AND COSINES

Finally then,

cos6 x dx = 1 (10 + 15 cos(2x) + 6 cos(4x) + cos(6x)) dx 32

1

15

6

1

= (10x + sin(2x) + sin(4x) + sin(6x)) + C.

32

2

4

6

Notice where we have divided by 2, 4, and 6 because we are integrating the cosine

of 2x, 4x, and 6x; we could have done this formally by setting u = 2x, etc., but it's

nice to learn to just divide by the constant in as simple a case as this.

If you choose only to learn the above algorithm (and not the extra material below) then in order to integrate something like

cos4(x) sin4(x) dx

you can replace sin4(x) with (1-cos2(x))2 = 1-2 cos2(x)+cos4(x) and then rewriting the integral as

cos4(x))(1 - 2 cos2(x) + cos4(x)) dx

= cos4(x) dx - 2 cos6(x) dx + cos8(x) dx.

You would then apply the algorithm above to solve each of the three integrals of powers of cosines of x.

2. Odd powers of cos x

For odd powers of the cosine function, use the same process as above, with a slight

change. The cosines of 0x, 2x, 4x, etc. become the cosines of 1x, 3x, 5x, etc. So for example, for cos5 x we write

1 11 121 1331 14641 1 5 10 10 5 1

and then

(20, 10, 2)

which translates to the identity

(2 cos(x))5 = 32 cos5(x) = 20 cos(x) + 10 cos(3x) + 2 cos(5x)

so

cos5 x =

1 (20 cos(x) + 10 cos(3x) + 2 cos(5x)).

32

NOTES ON HOW TO INTEGRATE EVEN POWERS OF SINES AND COSINES

3

3. The sine function

For the sine function, there are two changes, both being an "every other one" type of change.

First, every other term is negative. Second, odd powers of sin(x) get expanded in terms of sine functions, while even powers of sin(x) get expanded in terms of cosine functions.

Rather than elaborate further, we will simply list the identities for the first six powers of the cosine function and the first six powers of the sine function and let you learn the pattern by observation.

cos x = cos x

cos2 x = 1/2(1 + cos(2x))

cos3 x = 3 cos x + 1 cos(3x)

4

4

cos4 x = 3 + 1 cos(2x) + 1 cos(4x)

82

8

cos5 x = 5 cos x + 5 cos(3x) + 1 cos(5x)

8

16

16

cos6 x =

5

15 + cos(2x) +

3

cos(4x) +

1

cos(6x)

16 32

16

32

sin x = sin x

sin2 x = 1/2(1 - cos(2x))

sin3 x = 3 sin x - 1 sin(3x)

4

4

sin4 x = 3 - 1 cos(2x) + 1 cos(4x)

82

8

sin5 x = 5 sin x - 5 sin(3x) + 1 sin(5x)

8

16

16

sin6 x =

5

15 - cos(2x) +

3

cos(4x) -

1

cos(6x)

16 32

16

32

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NOTES ON HOW TO INTEGRATE EVEN POWERS OF SINES AND COSINES

4. Where the algorithm comes from

The ideas come from complex analysis. Suppose we have the equation

2 cos(x) = u + 1/u.

(1)

(This is true when u = cos(x)+i sin(x), if you know a little about complex numbers.) Then

(2 cos(x))2 = (u + 1/u)2 = u2 + 2 + 1/u2 = 2 + (u2 + 1/u2).

Now we also know that

(2 cos(x))2 = 4 cos2 x = 4(1/2 + 1/2 cos(2x)) = 2 + 2 cos(2x).

Combining this with the above we see that 2 + (u2 + 1/u2) = 2 + 2 cos(2x)

so that

u2 + 1/u2 = 2 cos(2x).

Taking it a step further, let's multiply this last equation again by (u + 1/u).

(u2 + 1/u2)(u + 1/u) = 2 cos(2x)2 cos(x).

Using an identity in your book for a product of cosines, (u3 + u + 1/u + 1/u3) = 2 cos(3x) + 2 cos(x).

Subtracting the original equation (1), we get u3 + 1/u3 = 2 cos(3x).

Continuing in this fashion, it turns out that for any positive integer k, uk + 1/uk = 2 cos(kx).

Now the work we did on a previous page with Pascal's triangle can be understood

as follows: (2 cos(x))6 = (u + 1/u)6

= 1u6 + 6u5u-1 + 15u4u-2 + 20u3u-3 + 15u2u-4 + 6u1u-5 + 1u-6

= u6 + 6u4 + 15u2 + 20 + 15u-2 + 6u-4 + u-6.

But notice that we can separate this into

u6

+

1/u6 +

6u4

+

6/u4 +

15u2 + 15/u2 +

20

NOTES ON HOW TO INTEGRATE EVEN POWERS OF SINES AND COSINES

5

which, in turn, equals 2 cos(6x) + 6(2 cos(4x)) + 15(2 cos(2x)) + 20,

leading to the same conclusion as before.

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