1.

a.

m is odd

b.

m is even and n is odd

c.

m and n are both even

2.

3.

a.

n is even

b.

n is odd

4.

a.

m is even

b.

n is odd

c.

n is even and m is odd

FORM 1:

CASE 1: m is odd

Since m is odd, let m = 2k + 1.

Then sin ^m x = sin ^{2k + 1} x = sin ^2k x sin x = (sin ² x) ^k sin x = (1 - cos ² x) ^k sin x.

Then you will let u = cos x and du = - sin x dx.

EXAMPLE 1:

SOLUTION:

Let u = cos x, then du = - sin x dx.

EXAMPLE 2:

SOLUTION:

Let u = cos x and du = - sin x dx.

CASE 2: n is odd

 Since n is odd, let n = 2k + 1.

Then cos ⁿ x = cos ^{2k + 1} x = cos ^2k x cos x = (cos ² x) ^k cos x = (1 - sin ² x) cos x.

Then you will let u = sin x and du = cos x dx.

 EXAMPLE 3:

 SOLUTION:

Let u = sin x, then du = cos x dx, and when x = 0, u = 0 and when x = p / 2, then u = 1.

 EXAMPLE 4:

 SOLUTION:

Let u = sin x, then du = cos x dx.

CASE 3: m and n even

When m and n are even, we will use the following trig identities.

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

FORM 2:

Here is how you can simplify tan ⁿ x.

You may have to use this substitution a couple of times in specific problems.

EXAMPLE 7:

SOLUTION:

Let u = tan x, then du = sec ² x dx.

 EXAMPLE 8:

 SOLUTION:

Let u = tan x, then du = sec ² x dx.

Here is how you can simplify cot ⁿ x.

Again, you may have to use this substitution a couple of times in specific problems.

EXAMPLE 9:

SOLUTION:

Let u = cot x, then du = -csc ² x dx.

FORM 3:

CASE 1: n is even

Here is the substitution for sec ⁿ x when n is even. Let n = 2k.

= (sec ² x) ^{k - 1} sec ² x = (tan ² x + 1) ^{k - 1} sec ² x

Here is the substitution for csc ⁿ x when n is even. Let n = 2k.

= (csc ² x) ^{k - 1} csc ² x = (cot ²x + 1) ^{k - 1} csc ² x

EXAMPLE 10:

SOLUTION:

Let u = tan x, then du = sec ² x dx.

EXAMPLE 11:

SOLUTION:

Let u = cot x, then du = - csc ² x dx.

CASE 2: n is odd

When n is odd, you will have to use integration by parts. If you need to review this method, refer back to supplemental notes 13.

EXAMPLE 12:

SOLUTION:

u = sec ³ x

dv = sec ² x dx

du = 3sec ² x(sec x tan x)dx

= 3sec ³ x tan x dx

v = tan x

Now we will have to use integration by parts on the new integral.

u = sec x

dv = sec ² x dx

du = sec x tan x dx

v = tan x

Now we sub the above ( ­ ) result back into the original problem.

EXAMPLE 13:

SOLUTION:

u = csc x

dv = csc ² x dx

du = -csc x cot x dx

v = - cot x

FORM 4:

CASE 1: m is even

Since m is even, let m = 2k, then tan ⁿ x sec ^m x will become the following.

tan ⁿ x sec ^m x = tan ⁿ x sec ^{2 k} x = tan ⁿ x sec ^{2 k - 2} x (sec ² x) = tan ⁿ x (sec ² x) ^{k - 1}sec ² x = tan ⁿ x (tan ² x + 1) ^{k - 1}sec ² x

Also, cot ⁿ x csc ^m x will be transformed into the following.

cot ⁿ x csc ^m x = cot ⁿ x csc ^{2 k} x = cot ⁿ x csc ^{2 k - 2} x (csc ² x) = cot ⁿ x (csc ² x) ^{k - 1}csc ² x = cot ⁿ x (cot ² x + 1) ^{k - 1}csc ² x

EXAMPLE 14:

SOLUTION:

Let u = cot x, then du = - csc ² x dx.

 EXAMPLE 15:

SOLUTION:

Let u = tan x, then du = sec ² x dx.

CASE 2: n is odd

Since n is odd, we will let n = 2k + 1 and substitute the following in for tan ⁿ x sec ^m x.

Here is the substitution for cot ⁿ x csc ^m x.

= ((csc ² x - 1) ^k csc ^{m -1} x ) csc x cot x

EXAMPLE 16:

SOLUTION:

Let u = sec x, then du = sec x tan x dx.

EXAMPLE 17:

SOLUTION A:

Let u = csc x, then du = -csc x cot x dx.

SOLUTION B:

Let u = cot x, then du = - csc ² x dx.

Which of these two solutions is correct? Both are. Use the trig identity cot ² x + 1 = csc ² x to prove this to yourself.

CASE 3: n is even and m is odd

Again, let n = 2k, then tan ⁿ x sec ^m x will be substituted with the following.

Also, cot ⁿ x csc ^m x will be substituted with the following.

After using these substitutions, you will use the method stated in FORM 3, CASE 2.

EXAMPLE 18:

SOLUTION:

We will now evaluate the integral of sec ³ x using integration by parts.

u = sec x

dv = sec ² x dx

du = sec x tan x dx

v = tan x