8.3 Trigonometric Substitution

Transcription

1 Trigonometric Substitution Three Basic Substitutions Recall the derivative formulas for the inverse trigonometric functions of sine, secant, tangent. () () (3) d d d ( sin x ) = ( tan x ) = +x ( sec x ) = x, x < x x, x > and the corresponding antiderivative formulas. (4) = x sin x+c, x < (5) (6) +x = tan x+c x x = sec x +C, x >

2 8.3 To motivate trigonometric substitution, we start with the integral in (4). Notice that if we let x = sinθ, then the quadratic expression under the radical can be rearranged so that the radical can be eliminated. x = sin θ = cos θ = cosθ Now since θ = sin x [ π/,π/], we must have cosθ 0 and absolute values are not needed in the last expression. Returning to (4) we have = x = cosθdθ sin θ cosθdθ cos θ cosθdθ = cosθ = dθ = θ+c = sin x+c

3 8.3 3 Example. Trig Substitution using x = sin θ Evaluate (7) x If < x < then we can let x = sinθ, π/ θ π/. Thus So that = cosθdθ cosθdθ x = cos θ = secθdθ = ln secθ +tanθ +C = ln + x x x +C = ln +x x +C Remark. Also, see remarks at the end of this section.

4 8.3 4 It is worthwhile to verify. d ( ln +x ) = d x (ln(+x) ln( x )) = +x ( x x = x x + x x = x ) In practice we will usually need to construct a reference triangle as shown below. The substitution (8) sinθ = x = x allows us to construct the (reference) triangle for θ. x θ x We usually want the substitution in (8) to be invertible. That is, (9) x = sinθ iff θ = sin x, π θ π

5 8.3 5 With minor modifications in the earlier discussion, we can handle more general quadratic expressions. In fact, we have the following. The substitution (0) () () u = asinθ iff θ = sin u a, π θ π u = atanθ iff θ = tan u a, π < θ < π u = asecθ iff θ = sec u a The last substitution () is valid whenever (3) and (4) 0 θ < π if u a π θ π if u a Remark. For secant substitutions, the author restricts the exercises so that the first case (3) applies. Also, in (0), we must have u/a.

6 8.3 6 Example. Which Trig Substitution is Right for You? Evaluate the integrals below. a) a 0 x, 0 a In section 8. we evaluated this integral using partial integration. This time we ll try trigonometric substitution. Let x = sinθ, 0 θ π/. The corresponding reference triangle is shown below. x θ x Then θ = sin x, = cosθdθ, θ(0) = 0 and θ(a) = sin a. It follows that a 0 sin x = a cos θdθ = = 0 ( θ + sinθ sin a 0 ) sin a +cosθ = (sin a+sin ( sin a ) cos ( sin a ) = (sin a+a a ) 0 dθ = (θ +sinθcosθ) sin a 0 As we saw in section 8..

7 8.3 7 b) +4x The integrand in this case should remind us of the reference triangle for the tangent function. Now let x = tanθ, 0 θ < π/. +4x x θ Now = (/)sec θdθ so that +4x = = secθ sec θdθ, (since θ 0) sec 3 θdθ = 4 secθtanθ+ 4 ln secθ+tanθ = 4 +4x (x)+ 4 ln +4x +x = x +4x + ( ) 4 ln +4x +x

8 8.3 8 c) 4x 9 x, x > 3 The integrand in this case should remind us of the reference triangle for the secant function. So we let x = 3secθ, 0 θ < π/. So x 4x 9 θ 3 So = 3 secθtanθdθ x = 9sec θ 4 4x 9 = 3tanθ

9 8.3 9 So that 4x 9 x = 3 ( 4 9 = )( ) 3 tanθ sec θ secθtanθdθ tan θcosθdθ sin θ = cosθ dθ cos θ = = cosθ dθ secθ cosθdθ...and we can drop the absolute values since x > 0. = ln secθ +tanθ sinθ+c = ln x 3 + 4x 9 3 4x 9 +C x ( ) x = ln 3 + 4x 9 4x 9 +C 3 x

10 8.3 0 An Advanced Substitution One occasionally encounters integrals involving rational expressions of sine and cosine (c.f., exercises 4-50 from the chapter 8 review). In many situations the substitution (5) z = tan x is often useful. (cosθ,sinθ) In the figure we have sketched a unit circle with center at O. By the Pythagorean theorem h = (+cosθ) +sin θ A θ/ h O θ cosθ sinθ = +cosθ +cos θ +sin θ = (+cosθ) It follows that sin θ = = sinθ sin θ = (+cosθ) (+cosθ) cos θ (+cosθ) Together with the sketch, we obtain the so-called Half-Angle formulas for the sine, cosine, and tangent functions. (6) (7) (8) sin θ = cosθ cos θ = +cosθ tan θ = sinθ +cosθ = cosθ sinθ

11 8.3 The identities (5), (6) and (7) together imply Rearranging produces z = sin(x/) cos(x/) = cosx +cosx cosx = z +z One can use the Pythagorean identities (or otherwise) to find Finally, (5) implies sinx = z +z x = tan z So that = dz +z (Also, see the discussion on page 494 of the text).

12 8.3 Example 3. Evaluate the following integral. +sinx Method : Multiplying by a Convenient Form of One, etc. +sinx = sinx +sinx sinx sinx = cos x = (sec x secx tanx) = tanx secx+c

13 8.3 3 Method : Using the z-substitution +sinx = = = + z +z dz +z +z dz (+z) = +z +C = z +z +tan(x/) +C =. = (+cosx) +sinx+cosx +C =. = tanx secx +C

14 8.3 4 Here s a more interesting example. Example 4. Evaluate 3cosx This time the z-substitution appears to be our only option. The substitutions yield cosx = z +z = dz +z +z 3cosx = dz 4z +z dz = z Hence, by partial fractions, we obtain ( = )dz z z + = ln z ln z + +C = z ln +C z + = tan(x/) ln +C tan(x/)+

15 8.3 5 Remark. Example is for illustrative purposes only. It is much better to handle the integral in (7) using partial fractions. Observe that x = x+ x (9) = (ln x+ ln x ) but without the restriction < x < as we had above. For example, it is clear that we can not evaluate the integral below using trig substitution (Why?). However, 5 x = 5 x+ 5 x = (ln x+ ln x ) 5 = [(ln6 ln4) (ln3 ln)] = ln We will discuss partial fractions in the next section.