Trig Integrals 
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 setup depends on
the integrand's category.
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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)
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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 halfangle formulas to change the integrand.
The halfangle 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 halfangle formula again to cos² 2x, so the integral becomes
Integrating this expression gives us
Notice that when we applied the ½angle formula the second time, the angle became 4x.
If we have a combo of even powers, we apply the halfangle formula to both trig functions.
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Example 4
Now we apply the halfangle 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:
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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
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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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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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1/ even power of both sine and cosine we use halfangle 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 
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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