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.

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

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.

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.

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∫