Intro:
Integrands with trig functions can be put in categories and then integrated with
standard techniques. Learn the categories, then follow the instructions. We may
have to call upon other techniques of integration such as integration by parts
to complete the question, but the initial approach and set-up depends on
the integrand's category.

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TABLE OF CONTENTS

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Category A: Powers of Sine and Cosine

Category B: Powers of Secant and Tangent

Category C: Powers of Cosecant and Cotangent

Category D: Products of sin (ax), cos (bx)

Practice & Solutions

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Category A: Powers of Sine and Cosine

1) Odd Powers of Sine and Cosine

When the integrand includes odd powers of sine, cosine or a combo of the two,
we rewrite the expression as the function to the first power times the remaining
even power of it. Then, using the Pythagorean identities,
(sin ² u + cos ² u = 1)
we change the even power of the sine or cosine function into the
other function so that sine or cosine to the first power is our du.

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Example 1

Now we set u = cos x and du = – sin x dx, so the integral becomes

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Example 2

Now we set u = sin x and du = cos x dx, so the integral becomes

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2) Even Powers of Sine and Cosine

When the integrand includes even powers of sine, cosine or a combo of the two,
we use the half-angle formulas to change the integrand.

The half-angle formulas are:

sin 2 x = ½ (1 – cos 2x)

cos 2 x = ½ (1 + cos 2x)

Note that the angle in the formula is twice the angle in the original expression,
so if we have to apply the formula more than once, we keep doubling the angle.

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Example 3

Now we apply the half-angle formula again to cos² 2x, so the integral becomes

Integrating this expression gives us

Notice that when we applied the ½-angle formula the second time, the angle became 4x.

If we have a combo of even powers, we apply the half-angle formula to both trig functions.

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Example 4

Now we apply the half-angle formula again to cos² 2x, so the integral becomes

Integrating this expression gives us

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Category B: Powers of Secant & Tangent

1) Even Powers of Secant

Since sec ² x is the derivative of the Tangent function, and since we can change
even powers of the secant function into tangent expressions with the identity
1 + tan² x = sec² x,
we grab a sec ² x to reserve as our du and then change the remaining even power of
secant into tangent expressions.

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Example 5

Now, we multiply and set u = tan x , so du = sec² x dx. The integral becomes

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2) Odd Powers of Tangent

Since sec x tan x is the derivative of the Secant function, and since we can change
even powers of the tangent function into secant expressions with tan² x = sec² x – 1,
we grab a sec x tan x to reserve as our du and then change the remaining even power of
tangent into secant expressions.

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Example 6

Now, we multiply and set u = sec x , so du = sec x tan x dx. The integral becomes

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Integrals with Secant and Tangent that don't fit these categories are handled with
other integration techniques such as substitution (change of variable)
or integration by parts. (see lesson titled "The Boomerang Integral")

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Category C: Powers of Cosecant & Cotangent

Even Powers of Cosecant

These work exactly the same way that powers of Secant and Tangent do.

Since csc² x is the derivative of cot x, and since we can change
even powers of cosecant into cotangent expressions with 1 + cot² x = csc² x,
we grab csc² x to reserve as our du and then change the remaining even power of
csc x into cotangent expressions. Then we substitute u = cot x and du = – csc² x dx.

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Odd Powers of Cotangent

Since csc x cot x is the derivative of csc x, and since we can change
even powers of the cotangent function into cosecant expressions with cot² x = csc² x – 1,
we grab a csc x cot x to reserve as our du and then change the remaining even power of
cotangent into cosecant expressions. Then we substitute u = csc x and du = – csc x cot x dx.

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Memory Tweak:

if a trig function name includes the "co" syllable, the derivative is negative.

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Category D: Integrands with Products of sin (ax) and cos (bx)

When we have integrands with a combo of sine and cosine functions multiplied,
we change the integrand using the product formulas which are:

1) sin u cos v = ½ [sin (u + v) + sin (u – v)]	2) cos u sin v = ½ [sin (u + v) – sin (u – v)]
3) cos u cos v = ½ [cos (u + v) + cos (u – v)]	4) sin u sin v = ½ [cos (u – v) – cos (u + v)]

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So, an integrand of the form sin 5x cos 2x becomes

using the first formula in the list.

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Example 7

since the antiderivative of sin u= ( – 1/du) cos u.

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Example 8

Here we used the 4th formula in the list.

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Practice

Integrate

1/ sin 4 x cos 2 x dx	2/ x sin 3 x 2 cos 2 x 2 dx
3/ sec x tan 3 x dx	4/ tan 3 x dx
5/ csc 4 x cot 3 x dx	6/ cos5x cos x dx

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Solutions

1/ even power of both sine and cosine we use half-angle formulas: (sin 2 x) 2 cos 2 x dx= [ ½(1 – cos 2x) ] 2 ½(1 + cos 2 x) dx = 1/8 [ 1 – 2cos 2x + cos 2 2x ](1 + cos 2 x) dx = 1/8 (cos 3 2 x – cos 2 2x – cos 2x) dx use odd power approach on cos 3 2 x use even power approach on cos 2 2x	2/ x sin 3 x 2 cos 2 x 2 dx First substitute w = x² and dw = 2x dx The integral now becomes ½ sin 3 w cos 2 w dw use odd power approach on sin 3 w ½ (1 – cos 2 w) cos 2 w sin w dw after "unsubstitution", we get
3/ tan 2 x sec x tan x dx = (u² – 1) du	4/ tan x (sec 2 x – 1)dx = tan x sec 2 x dx – tan x dx =
5/ this is an even power of csc x csc 2 x csc 2 x cot 3 x dx = cot 3 x (cot 2 x – 1) csc 2 x dx = – ( u 5 – u 3 ) du = ¼ cot 4 x – 1/6 cot 6 x + C	6/ using the 3rd product formula ½ (cos 6x + cos 4x ) dx =

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(Calculus II MathRoom Index)

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