recap with trig techniques - lexington public...
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![Page 1: Recap with Trig Techniques - Lexington Public Schoolslhsteacher.lexingtonma.org/haupt/bc_calc/RecapwithTrigTechniques.… · ∫cos2(bx)dx by using the power reducing identities derived](https://reader034.vdocuments.net/reader034/viewer/2022052320/605a1cd42b9ec424ca702b15/html5/thumbnails/1.jpg)
Honors AP Calculus BC Name:______________________________ Trig Integration Techniques 13 December 2013
Integration Techniques • Antidifferentiation • Substitutiion (antidifferentiation of the Chain rule) • Integration by Parts (antidifferentiation of the Product rule)
o straightforward o repeated (tabular) o circular
• Trigonometric Substitution (for
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a2 − x 2 ,
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x 2 − a2 ,
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a2 + x 2 forms – see page 2) • Separation of Variables (lesson on December 16th) • Partial Fraction Decomposition (when
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1(x−a )(x−b ) is involved – lesson on December 18h)
• Powers of Trig Functions (see pages 3-5) • Laplace Transforms (after Christmas break)
Solving Differential Equations (specifically Initial Value Problems) • Graphically (Slope Fields – draw possible solution from initial point) • Numerically (Euler’s Method – calculate approximate values from initial point) • Analytically (using integration techniques, initial value determines C)
Analytically, we can currently solve differential equations of the form:
•
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dydx
= ax and
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dydx
= af (x) [antidifferentiation]
•
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dydx
= a ʹ′ f (g(x))⋅ ʹ′ g (x) [substitution]
•
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dydx
= af (x)⋅ ʹ′ g (x) [integration by parts]
but we cannot yet solve differential equations of the form:
•
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dydx
= ay and
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dydx
= af (x)⋅ g(y) [separation of variables]
•
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dydx
= f (x) + g(y)
•
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d2ydx 2
= a ʹ′ y + by (more often written as
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f (x) = ʹ′ ʹ′ y + 2b ʹ′ y + cy )
The last two forms require a technique called Laplace Transforms which is not in the book but is really interesting and useful.
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Trig Integration Techniques page 2
Trig Substitution In finding the area of a circle, we usually employ a geometric method rather than trying to integrate
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r2 − x 2∫ dx . However, this form shows up in situations where a purely geometric method is not available. In this case we will use another form of substitution called trigonometric substitution. Trigonometric Substitution is based on the relationship of the sides in a right triangle. We are able to use another variable to simplify the integral.
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sinθ =xa
x = a sinθ
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a2 − x 2 = a2 − (asinθ)2
= a2(1− sin2θ )= acosθ
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tanθ =xa
x = a tanθ
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a2 + x 2 = a2 + (atanθ )2
= a2(1+ tan2θ)= asecθ
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secθ =xa
x = a secθ
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x 2 − a2 = (asecθ)2 − a2
= a2(sec2θ −1)= atanθ
Let us look at the example of the circle:
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9 − x 2∫ dx . In this case, 9 – x2 is a2 – x2 so a = 3,
x = 3sinθ and
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9 − x 2 = 3cosθ . We also need to keep in mind that dx = 3cosθ dθ.
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9 − x 2∫ dx = 3cosθ∫ ⋅ 3cosθdθ = 9 cos2θdθ = 9 1− cos(2θ( )∫∫ dθ = 9 12θ −
14 sin(2θ)( ) +C
Using x = 3sinθ to solve for θ and substituting back in gives
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9 12 sin
−1 x3( ) − 1
4 sin 2sin−1 x
3( )( )( ) +C .
Not every problem has such a messy answer.
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9 − x 2
x 2∫ dx looks worse as a problem but see
what happens when we do the trigonometric substitution used in the last example.
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9 − x 2
x 2∫ dx =3cosθ9sin2θ∫ ⋅ 3cosθdθ = cot2θdθ = csc2θ −1( )∫∫ dθ = cotθ −θ( ) +C = cot sin−1 x
3( )( ) − sin−1 x3( ) +C
Problems in the book: p.338, 340/47 – 52 and 81 – 84
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Trig Integration Techniques page 3
Powers of Trig Functions We have done the following trigonometric integration problems:
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sin(bx)dx = − 1b cos(bx) +C∫
•
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cos(bx)dx = 1b sin(bx) +C∫
•
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sin(bx)cos(bx)dx = 12b sin
2(bx) +C = − 12b cos
2(bx) +C∫ (use substitution with u = sine or cosine) •
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sec2(bx)dx = 1b tan(bx) +C∫
•
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sec(bx)tan(bx)dx = 1b sec(bx) +C∫
•
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tan(bx)dx = − 1b lncos(bx) +C = 1
b lnsec(bx) +C∫ (rewrite tan(bx) in terms of sine and cosine, then use substitution)
We have also dealt with
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sin2(bx)dx∫ and
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cos2(bx)dx∫ by using the power reducing identities derived last year:
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sin2 x = 12 1− cos(2x)( ) and
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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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sec3(bx)dx∫ . Below are examples of techniques available to address such problems. All of the examples have an integrand with sin3(x) and an increasing number of cosines. Example 1:
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sin3(x)dx∫
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sin3(x)dx∫ = 1− cos2 x( )∫ sin(x)dx = sin(x)dx − u2du∫∫ Example 2:
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sin3(x)cos(x)dx∫
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sin3(x)cos(x)dx∫ = u3du∫ Example 3:
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sin3(x)cos2(x)dx∫
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sin3(x)cos2(x)dx∫ = sin(x) 1− cos2 x( )cos2(x)dx = cos2 x − cos4 x( )sin(x)dx = u4 − u2( )du∫∫∫ Example 4:
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sin3(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
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sin3(x)cos3(x)dx∫ = sin(x)cos(x)( )3dx∫ = 12 sin(2x)( )∫
3dx and finish like example 1.
These examples lead us to some strategies for evaluating
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sinm (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
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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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Trig Integration Techniques page 4 There are similar examples and strategies when working with tangent and secant but we need to be able to find the indefinite integral
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sec(x)dx∫ first.
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sec(x)dx∫ = sec(x) sec(x) + tan(x)sec(x) + tan(x)
dx∫ =sec2(x) + sec(x)tan(x)
sec(x) + tan(x)dx∫ .
At this point we substitute u = sec(x) + tan(x) and du = (sec(x)tan(x) + sec2(x))dx. so
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sec(x)dx∫ =duu
= ln sec(x) + tan(x) +C∫
Example 5:
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tan3(x)dx∫
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tan3(x)dx∫ = tan(x) sec2 x −1( )∫ dx = u⋅ du −∫ tan(x)dx∫
Example 6:
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tan3(x)sec(x)dx∫
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tan3(x)sec(x)dx∫ = tan2(x)tan(x)sec(x)dx∫ = sec2 x −1( ) tan(x)sec(x)dx = u2du − tan(x)sec(x)dx∫∫∫
Example 7:
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tan3(x)sec2(x)dx∫
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tan3(x)sec2(x)dx∫ = u3du∫ Example 8:
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tan3(x)sec3(x)dx∫
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tan3(x)sec3(x)dx∫ = sec2 x −1( )sec2(x) tan(x)sec(x)dx( ) = u2 −1( )u2du∫∫
Example 9:
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sec3(x)dx∫ Your first thought might be to rewrite this as
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tan2(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) and dv = sec2(x)dx.
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sec3(x)dx = sec(x)tan(x) −∫ sec(x)tan2(x)dx ∫= sec(x)tan(x) − sec(x) sec2(x) −1( )dx ∫= sec(x)tan(x) − sec3(x)dx∫ + sec(x)dx∫
Strategies for evaluating
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tanm (x)secn (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)
to express the remaining factors in terms of tangent. (c) Other cases are not as clear cut and will require trying strategies like integration by parts
with the identities.
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Trig Integration Techniques page 5 Final notes:
(a) cosecant and cotangent follow similar strategies to secant and tangent. Be careful with the negatives.
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csc2(x)dx = −cot(x) +C∫
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csc(x)cot(x)dx = −csc(x) +C∫
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cot(x)dx = ln sin(x) +C = −lncsc(x) +C∫
(b) If you must evaluate
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sin(mx)cos(nx)dx∫ then it is helpful to employ the identities:
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sinAcosB = 12 sin(A − B) + sin(A + B)[ ]
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sinAsinB = 12 cos(A − B) − cos(A + B)[ ]
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cosAcosB = 12 cos(A − B) + cos(A + B)[ ]
Problems 1.
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cos3(x)dx∫ 2.
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sin5(x)cos2(x)dx∫ 3.
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sin4 (x)dx∫ 4.
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sin2(πx)cos4 (πx)dx∫ 5.
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tan6(x)sec4 (x)dx∫ 6.
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tan3(2x)sec5(2x)dx∫ 7.
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cot3(x)csc3(x)dx∫ 8.
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cot4 (x)csc6(x)dx∫ 9.
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sin(8x)cos(5x)dx∫ 10.
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cos(πx)cos(4πx)dx∫