Recap of Ch 6 so far
page 2
Powers of Trig Functions We have done the following trigonometric integration problems:
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1
b
1
b
1
2b
2
2
2
1
2b
1
b
1
b
1
b
1
b
then use substitution)
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∫ sin(bx)dx = − cos(bx) + C
∫ cos(bx)dx = sin(bx) + C
∫ sin(bx)cos(bx)dx = sin (bx) + C = − cos (bx) + C (use substitution with u = sine or cosine)
∫ sec (bx)dx = tan(bx) + C
∫ sec(bx)tan(bx)dx = sec(bx) + C
∫ tan(bx)dx = − ln cos(bx) + C = ln sec(bx) + C (rewrite tan(bx) in terms of sine and cosine,
We have also dealt with
∫ sin (bx)dx and ∫ cos (bx)dx by using the power reducing identities
derived last year: sin 2 x =
2
1
2
2
(1 − cos(2x)) and cos2 x = 12 (1+ cos(2x)) .
However, we have struggled a bit when the powers on the trig functions got higher, for example
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. Below are examples of techniques available to address such problems. All of the
∫ sec 3 (bx)dx
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examples have an integrand with sin3(x) and an increasing number of cosines.
∫ sin (x)dx
∫ sin (x)dx = ∫ (1 − cos x ) sin(x)dx = ∫ sin(x)dx − ∫ u du
Example 1:
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3
3
2
2
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Example
2: ∫ sin 3 (x)cos(x)dx
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∫ sin3 (x)cos(x)dx = ∫ u3du
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Example
3: ∫ sin 3 (x)cos2 (x)dx
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∫ sin3 (x)cos2 (x)dx = ∫ sin(x)(1 − cos2 x ) cos2 (x)dx =
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Example
4: ∫ sin 3 (x)cos3 (x)dx
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∫ sin3 (x)cos3 (x)dx = ∫ sin(x)(1 − cos2 x ) cos3 (x)dx and finish like example 3 or
3
∫ sin (x)cos (x)dx = ∫ (sin(x)cos(x)) dx = ∫ (
3
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3
1
2
∫ (cos
2
x − cos4 x ) sin(x)dx =
∫ (u
4
− u 2 ) du
3
sin(2x)) dx and finish like example 1.
These examples lead us to some strategies for evaluating ∫ sin m (x)cosn (x)dx
(a) If the power of the sine is odd (m is odd), save one sine factor and use sin2(x) = 1 – cos2(x)
to express the remaining factors in terms of cosine
(b) If the power of the cosine is odd (n is odd), save one cosine factor and use cos2(x) = 1 – sin2(x)
to express the remaining factors in terms€of sine
(c) If the powers are both odd, either technique can be used or the half-angle identity
sin(x)cos(x) = 12 sin(2x) can be used (see example 4 above).
(d) If the powers of both sine and cosine are even, use the power-reducing identities
mentioned above and then revisit strategies a, b, or c.
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Recap of Ch 6 so far
page 3
There are similar examples and strategies when working with tangent and secant but we need to
be able to find the indefinite integral ∫ sec(x)dx first.
sec(x) + tan(x)
∫ sec(x)dx = ∫ sec(x) sec(x) + tan(x) dx = ∫
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sec 2 (x) + sec(x)tan(x)
dx .
sec(x) + tan(x)
At this point we substitute u = sec(x) + tan(x) and du = (sec(x)tan(x) + sec2(x))dx. so
du
∫ sec(x)dx = ∫ u = ln sec(x) + tan(x) + C
∫ tan (x)dx
∫ tan (x)dx = ∫ tan(x)(sec
Example 5:
3
3
2
x −1)dx =
∫ u⋅ du − ∫ tan(x)dx
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Example
6: ∫ tan 3 (x)sec(x)dx
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∫ tan3 (x)sec(x)dx = ∫ tan2 (x)tan(x)sec(x)dx =
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Example
7: ∫ tan 3 (x)sec 2 (x)dx
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∫ tan3 (x)sec 2 (x)dx = ∫ u3du
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Example
8: ∫ tan 3 (x)sec 3 (x)dx
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∫ tan3 (x)sec 3 (x)dx = ∫ (sec 2 x −1) sec 2 (x)(tan(x)sec(x)dx ) =
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∫ (sec
2
x −1) tan(x)sec(x)dx =
∫ (u
2
∫ u du − ∫ tan(x)sec(x)dx
2
−1) u 2 du
Example
9: ∫ sec 3 (x)dx
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Your first thought might be to rewrite this as ∫ (tan 2 (x) +1) sec(x)dx but this will lead to a
circular situation that ends with 0 = 0. It is better to use integration by parts with u = sec(x)
2
€and dv = sec (x)dx.
(x)dx
∫ sec (x)dx = sec(x)tan(x) − ∫ sec(x)tan
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= sec(x)tan(x) − ∫ sec(x)(sec (x) −1) dx
= sec(x)tan(x) − ∫ sec (x)dx + ∫ sec(x)dx
3
2
2
3
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Strategies for evaluating ∫ tan m (x)sec n (x)dx
(a) If the power of tangent is odd (m is odd), save a factor of sec(x)tan(x) and use tan2(x) = sec2(x) – 1
to express the remaining factors in terms of secant.
(b) If the power of secant is even (n is even), save a factor of sec2(x) and use sec2(x) = 1 + tan2(x)
€ the remaining factors in terms of tangent.
to express
(c) Other cases are not as clear cut and will require trying strategies like integration by parts
with the identities.
Recap of Ch 6 so far
page 4
Final notes:
(a) cosecant and cotangent follow similar strategies to secant and tangent. Be careful with
the negatives.
∫ csc 2 (x)dx = −cot(x) + C
∫ csc(x)cot(x)dx = −csc(x) + C
∫ cot(x)dx = ln sin(x) + C = −ln csc(x) + C
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(b) If you must evaluate
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Problems 1.
2.
3.
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5.
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6.
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7.
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8.
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9.
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10.
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∫ sin(mx)cos(nx)dx then it is helpful to employ the identities:
sin Acos B = [sin(A − B) + sin(A + B)]
sin Asin B = [cos(A − B) − cos(A + B)]
cos Acos B €
= [cos(A − B) + cos(A + B)]
1
2
1
2
1
2
∫ cos (x)dx
∫ sin (x)cos (x)dx
∫ sin (x)dx
∫ sin (πx)cos (πx)dx
∫ tan (x)sec (x)dx
∫ tan (2x)sec (2x)dx
∫ cot (x)csc (x)dx
∫ cot (x)csc (x)dx
∫ sin(8x)cos(5x)dx
∫ cos(πx)cos(4πx)dx
3
5
2
4
2
6
4
4
3
5
3
3
4
6