Lecture 25 Notes: Integration by Inverse Substitution

18.01 Calculus Jason Starr Fall 2005 Lecture 25. November 17, 2005 Homework. Problem Set 7 Part I: (a)–(e) Practice Problems. Course Reader: 5D­2, 5D­6, 5D­7, 5D­10, 5D­14 1. Inverse hyperbolic functions. There are a few other useful formulas for hyperbolic functions; for instance, the analogues of the angle­addition formulas, sinh(s + t) = sinh(s) cosh(t) + cosh(s) sinh(t), cosh(s + t) = cosh(s) cosh(t) + sinh(s) sinh(t). These imply the double­angle formulas, sinh(2t) = 2 sinh(t) cosh(t), cosh(2t) = cosh2 (t) + sinh2 (t) = 2 cosh2 (t) − 1 = 2 sinh2 (t) + 1. From these follow the analogues of the half­angle formulas, 1 sinh2 (t/2) = (cosh(t) − 1), 2 1 cosh2 (t/2) = (cosh(t) + 1). 2 A beautiful feature of hyperbolic functions is that their inverse functions can be expressed in terms of simpler functions. The inverse function sinh−1 (x) of sinh(x) is defined on the whole real line. By definition, sinh−1 (x) = y if and only if sinh(y) = x. This second equation can be written out as, 1 y (e − e−y ) = x. 2 Substituting z = ey gives, 1 (z − z −1 ) = x. 2 Multiplying both sides by 2z gives, z 2 − 1 = 2xz ⇔ z 2 − 2xz − 1 = 0. Completing the square gives, (z − x)2 = x2 + 1. Taking square roots gives, z =x± √ x2 + 1. 18.01 Calculus Jason Starr Fall 2005 Since z equals ey , z is positive. Thus, the correct square root is, √ z = x + x2 + 1. Finally this gives, y = ln(z) = ln(x + √ x2 + 1). Therefore, the formula for the inverse hyperbolic sine is, sinh−1 (x) = ln(x + √ x2 + 1). The same type of argument also gives, cosh−1 (x) = ln(x + √ x2 − 1), x ≥ 1, and tanh−1 (x) = (1/2) ln((1 + x)/(1 − x)), −1 < x < 1. 2. Derivatives of the inverse hyperbolic functions. By the same methods used to compute the derivatives of inverse trigonometric functions, the derivatives of the inverse hyperbolic functions are, du d sinh−1 (u) = √ , 1 + u2 du d cosh−1 (u) = √ , u ≥ 1, u2 − 1 du d tanh−1 (u) = , −1 < u < 1. 1 − u2 These can also be computed using the formulas for the inverse functions. 3. Inverse substitution. The derivatives of inverse trigonometric and inverse hyperbolic functions √ � allow us to compute more antiderivatives than before, e.g., dx/( x2 − 1) equals cosh−1 (x) + C. Essentially this comes down to making a direct substitution of an inverse function, e.g., u = cosh−1 (x). However, this is logically equivalent to making an inverse substition, x = cosh(u). When the integrand is more complicated, inverse substitution is usually simpler and faster than direct substitution of an inverse function. Example. Compute the following antiderivative, � √ a2 − x2 dx. This is not quite the derivative of an inverse function above. However, it is clear that inverse substituting x = a sin(θ) will simplify the integrand, because a2 − x2 = a2 − (a sin(θ))2 = a2 (1 − sin2 (θ)) = a2 cos2 (θ). 18.01 Calculus Jason Starr Fall 2005 Thus we have, � � √ � � � x = a sin(θ), 2 2 2 2 2 a − x dx, ,⇒ a cos (θ)(a cos(θ)dθ) = a cos2 (θ)dθ. dx = a cos(θ)dθ Using the half­angle formula, this becomes, � � � 1 1 θ 1 2 2 + sin(2θ) + C. a + cos(2θ)dθ = a 2 2 2 4 Using the double­angle formula and back­substituting gives, � √ √ a2 − x2 dx = (1/2)(a2 sin−1 (x/a) + x a2 − x2 ) + C. 4. Three different kinds of integrals, three kinds of inverse substitution. The type of antiderivative where inverse substitution is most successful has the form, √ � F (x, Ax2 + Bx + C) √ dx, G(x, Ax2 + Bx + C) where A, B and C are constants, and F (x, y) and G(x, y) are polynomial functions in the two argu­ ments. Inverse substitution together with partial fractions solves all such antiderivative problems. The first step is to complete the square of the expression Ax2 + Bx + C. This gives, � �2 B B 2 − 4AC 2 − Ax + Bx + C = A x + . 2A 4A In particular, making the substition, u=x+ B , du = dx, 2A transforms the quadratic into one of 3 possible types, β 2 u2 + α2 , β 2 u2 − α2 , −β 2 u2 + α2 , where, � β= � |A|, α = |B 2 − 4AC | . |4A| Defining a = α/β, finally the integral is transformed to one of 3 possible types, √ � FI (u, a2 − u2 ) √ Type I: du, GI (u, a2 − u2 ) 18.01 Calculus Jason Starr Fall 2005 � √ FII (u, u2 − a2 ) √ du, GII (u, u2 − a2 ) � √ FIII (u, a2 + u2 ) √ du. GIII (u, a2 + u2 ) Type II: and Type III: For each of these types, there are 3 possible inverse substitutions: trigonometric, hyperbolic and rational. A flow chart of the 9 possible outcomes will be posted on the course webpage. Here are a couple of examples. In each example, the inverse rational substitution is given, although it was only briefly discussed in lecture. Example. Compute the following antiderivative, � x2 √ dx. a2 − x 2 The trigonometric inverse substition is, x = a sin(θ), dx = a cos(θ)dθ. The new antiderivative is, a2 sin2 (θ) � � a2 − a2 sin2 (θ) (a cos(θ)dθ). Simplifying gives, � a2 sin2 (θ)dθ. This can be simplified using the half­angle formula, � 1 1 2 − cos(2θ)dθ. a 2 2 This is easily seen to be, a 2 � � θ 1 − sin(2θ) + C. 2 4 Using the double­angle formula and back­substituting, � √ x2 √ dx = (1/2)(a2 sin−2 (x/a) − x a2 − x2 ) + C. a2 − x 2 Alternatively, the hyperbolic inverse substitution is, x = a tanh(t), dx = asech2 (t)dt. 18.01 Calculus Jason Starr Fall 2005 The new antiderivative is, a2 tanh2 (t) � asech2 (t)dt. a2 sech2 (t) � Simplifying gives, a 2 � 2 tanh (t)sech(t)dt = a 2 � sinh2 (t) dt. cosh3 (t) This can be simplified a bit by multiplying numerator and denominator by cosh(t) and then ex­ pressing in terms of sinh(t) as much as possible, � � sinh2 (t) sinh2 (t) 2 2 a cosh(t)dt = a cosh(t)dt. cosh4 (t) (1 + sinh2 (t))2 Make the substitution u = sinh(t), du = cosh(t)dt to get, � u2 2 a du. (1 + u2 )2 This can be rewritten as, a 2 � 1 du − a2 2 1+u � 1 du. (1 + u2 )2 The first of these terms is just a2 tan−1 (u). However, the second term requires another inverse substitution. All in all, this is not a very efficient approach. Finally, the rational inverse substitution is, x=a 2t 2(1 − t2 ) , dx = a dt. 1 + t2 (1 + t2 )2 The point is that, a2 − x 2 = a 2 (1 − t2 )2 . (1 + t2 )2 Thus the new antiderivative is, � 1 + t2 2a(1 − t2 ) 4a2 t2 dt. (1 + t2 )2 a(1 − t2 ) (1 + t2 )2 This simplifies to, 2 8a � t2 dt = 8a2 (1 + t2 )3 � 1 dt − 8a2 (1 + t2 )2 � 1 dt. (1 + t2 )3 Notice, these two integrals are the same type that occurred with inverse hyperbolic substitution. But they came up more quickly: rational inverse substitution is more efficient than inverse hyper­ bolic substitution for this problem. However, both require a further inverse trigonometric substi­ tution. So inverse trigonometric substitution is the most efficient for this problem. 18.01 Calculus Jason Starr Fall 2005 Example. Compute the following antiderivative, � x2 √ dx. x 2 − a2 The trigonometric inverse substitution is, x = a sec(θ), dx = a sec(θ) tan(θ)dθ. The new antiderivative is, a2 sec2 (θ) � � a2 sec2 (θ) − a2 a sec(θ) tan(θ)dθ. Because sec2 (θ) − 1 equals tan2 (θ), simplifying gives, � � 2 3 2 a sec (θ)dθ = a 1 dθ. cos3 (θ) This can be simplifed by multiplying numerator and denominator by cos(θ) and then expressing in terms of sin(θ) as much as possible, � � 1 1 2 2 cos(θ)dθ = a cos(θ)dθ. a 4 cos (θ) (1 − sin2 (θ))2 Make the substitution u = sin(θ), du = cos(θ)dθ) to get, � 1 2 a du. (1 − u2 )2 This can be computed using partial fractions (not yet discussed). Alternatively, the hyperbolic inverse substitution is, x = a cosh(t), dx = a sinh(t)dt. The new antiderivative is, � a2 cosh2 (t) � a sinh(t)dt. a2 cosh2 (t) − a2 Since cosh2 (t) − 1 equals sinh2 (t), simplifying gives, � 2 a cosh2 (t)dt. This can be simplified using the analogue of the half­angle formula, � 1 1 2 a + cosh(2t)dt. 2 2 18.01 Calculus Jason Starr Fall 2005 This is easily seen to be, a 2 � � t 1 − sinh(2t) + C. 2 4 Using the double­angle formula and back­substituting, � � √ x2 1� 2 √ dx = a cosh−1 (x/a) − x x2 − a2 . 2 x 2 − a2 Using the formula for cosh−1 (x/a), this becomes, � √ √ x2 √ dx = (1/2)(a2 ln(x + x2 − a2 ) − x x2 − a2 ) + C. x 2 − a2 Finally, the rational substitution is, 1 + t2 −(1 − t2 ) , dx = a dt. 2t 2t2 x=a The point is that, a2 − x 2 = a 2 Thus the new antiderivative is, � (1 − t2 )2 . (2t)2 2t −a(1 − t2 ) a2 (1 + t2 )2 dt. 4t2 a(1 − t2 ) 2t2 This simplifies to, a2 − 4 � (1 + t2 )2 a2 dt = − t3 4 � 1 2 + + tdt. 3 t t This evaluates to, a2 − 4 � −1 t + 2 ln(t) + 2 2 2t � + C. This is clearly the easiest of the 3 methods for computing the antiderivative, for this problem. However, there still remains the formidable problem of solving for t = t(x), back­substituting, and simplifying the resulting expression. All in all, inverse hyperbolic substitution is the most efficient for this problem.