LECTURE 33 Integration via a Change of Variables In Lectures 12 and 13 where we developed a general technique for computing derivatives that was based on two different kinds of results. First, we had a table that listed the derivatives of a relatively small set of elementary functions d n n 1 d d 1 1 x = nx sin (x) = cos (x) sin− (x) = dx − dx dx √1 x2 d eλx = λeλx d cos (x) = sin (x) d cos 1 (x) = − 1 dx dx dx − √1 x2 d 1 d − 2 d 1 − 1 − ln x = tan (x) = sec (x) tan− (x) = 2 dx | | x dx dx 1+x d ax = ln a ax d cot (x) = csc2 (x) d cot 1 (x) = 1 dx | | dx − dx − − 1+x2 Secondly, we had a short list of rules by which the derivatives of more complicated functions could be reduced to the computations involving the derivatives listed above: d df dg The Sum Rule: (f + g) = + • dx dx dx d df The Constant Multiplier Rule: (cf) = c if c is a constant • dx dx d df dg The Product Rule: (fg) = g + f • dx dx dx d f f g fg The Quotient Rule: = − , where f := df , g := dg • dx g g2 dx dx d df dg The Chain Rule: (f (g (x))) = • dx du dx u=g(x) Similarly, we have developed rules for computing the integrals (a.k.a. anti-derivatives) of functions by knowing a few basic anti-derivatives and a few rules for computing the anti-derivatives of more complicated functions built from our set of basic function. In fact, our Integral Table can be obtained directly from the Table of Derivatives by reinterpreing the function being differentiated as the anti-derivative of its derivative: n 1 n+1 1 1 x dx = x + C cos (x) dx = sin (x) + C dx = sin− (x) + C n+1 √1 x2 λx 1 λx 1− 1 e dx = λ e + C sin (x) = cos (x) + C 1+x2 dx = tan− (x) + C 1 2 − x dx = ln x + C sec (x) dx = tan (x) + C x a|x | 2 a dx = ln a + C csc (x) dx = cot (x) + C | | − However, our corresponding list of Integration Rules (at least right now) is considerably shorter: The Sum Rule: (f (x) + g (x)) dx = f (x) dx + g (x) dx • The Constant Multiplier Rule: cf (x) dx = c f (x) dx • Today, we shall add a new rule of integration, the Change of Variable Rule, which is basically the analog of the Chain Rule for differentiation. (This Change of Variable Rule is also called the Substitution Rule.) 140 33. INTEGRATION VIA A CHANGE OF VARIABLES 141 Theorem 33.1 (The Change of Variables Rule). Suppose the integrand of an integral is of the the form g (h (x)) h (x) (with g (x) continuous and h (x) differentiable), then u=h(x) (1) g (h (x)) h (x) dx = g (u) du + C Proof. This is very easy. To compute g (h (x)) h (x) dx we need to find an anti-derivative of g (h (x)) h (x). Suppose G is an anti-derivative of g (x); then g (u) du G (x) + C ≡ and u=h(x) (2) g (u) du = G (h (x)) + C But, now if we differentiate the right-hand-side of (2), applying the Chain Rule, we get G (h (x)) h (x) g (h (x)) h (x) ≡ u=h(x) This shows that g (u) du is an anti-derivative of g (h (x)) h (x). Thus u=h(x) g (h (x)) h (x) dx = anti-derivative of g (h (x)) h (x) = g (u) du Example 33.2. Compute 2x sin x2 + 1 dx using the Change of Variables Rule. The key thing here is to figure out how to view the integrand sin x2 + 1 dx 2 as being of the form g (h (x)) h (x). Suppose we take h (x) = x + 1 and g (u) = sin (u). Then 2 g (h (x)) h (x) = sin x + 1 (2x) which is exactly our integrand. We can now apply the Change of Variables Theorem 2 u=x +1 2 2x sin x2 + 1 dx = sin (u) du + C = cos (u) u=x +1 + C = cos x2 + 1 + C − | − Remark 33.3. A convenient way to remember the Change of Variable Rule is via a mneumonic arising from the notion of differentials. Recall that if u is a function of x, its differential is du = u (x) dx If u = h (x), then means du = h (x) dx and g (u) du = g (h (x)) h (x) dx. 33. INTEGRATION VIA A CHANGE OF VARIABLES 142 or (3) g (h (x)) h (x) dx = g (u) du which is how the text writes the Change of Variable Rule. It is important to remember though, that compute the right hand side of by first finding anti-derivative G (u) of g (u), and then substituting h (x) for u in the final result G (h (x)). ( Presumably, this is why the text calls this method the Substitution Rule). Example 33.4. Compute 2 xex dx We want to view the integrand as being of the form g (u) du. It it natural to try g (u) = eu u (x) = x2 = du = 2xdx ⇒ However, the integrand is not quite 2 g (u) du = 2xex — we’re off by a factor of 2. On the other hand, we can write 2 1 2 1 2 d 1 1 1 2 xex dx = 2xex dx = ex x2 dx = eudu = eu = ex 2 2 dx 2 2 2 Example 33.5. Try to compute 2 ex dx Here we might try the same thing g (u) = eu u (x) = x2 = du = 2xdx ⇒ Again we have 2 g (u) du = ex (2x) dx but we have a more serious problem. We can not get rid of the extra factor of (2x) by simply mulitplying by a constant factor. In fact, the integral 2 ex dx is one of the classical examples of a integral for which can not be computed in terms of elementary functions. (It is perfectly well defined as an integral, it’s just that the answer can’t be expressed in terms of elementary functions.) In the next couple of examples I’ll try to demonstrate the sort of thinking process that one needs to use in order to successfully apply the Change of Variables Method. Example 33.6. Compute esin(2x+1) cos (2x + 1) dx 33. INTEGRATION VIA A CHANGE OF VARIABLES 143 The first thing to look for is the most complicated factor in the integrand; more precisely, the factor • that is the most complicated composition of functions. In the case at hand, it is the factor esin(2x+1) which is a composition where x 2x + 1 sin (2x + 1) esin(2x+1). −→ −→ −→ The last operation applied is sin (2x + 1) esin(2x+1) and this suggests that we should try thinking of this factor as eu with u = sin (2x + 1).−→ Now d u = sin (2x + 1) = du = sin (2x + 1) dx = (cos (2x + 1) 2) dx = 2 cos (2x + 1) dx ⇒ dx Let us think of this last equation as 1 dx = du 2 cos (2x + 1) Inside the orginal integral we now make the substitutions esin(2x+1) eu −→ 1 dx du −→ 2 cos (2x + 1) to get 1 1 1 esin(2x+1) cos (2x + 1) dx = eu cos (2x + 1) du = eudu = eu + C 2 cos (2x + 1) 2 2 The final step is to remember that u = sin (2x + 1). After making this last substitution we get 1 esin(2x+1) cos (2x + 1) dx = esin(2x+1) + C 2 Example 33.7. Compute x dx √1 4x2 − The integrand factors as • 1 x · √1 4x2 − The second factor is a composition of the form 1 1/2 x 1 4x2 = 1 4x2 − −→ − −→ √1 4x2 − − which suggests that we try 1/2 g (u) = u− 1/2 2 since u u− corresponds to the last function applied and u = 1 4x , since that is the argument of the last−→ function to applied. We have − u = 1 4x2 = du = 1 4x2 dx = 8x dx − ⇒ − − or 1 dx = du −8x We now make the substitutions 1 1/2 u− √1 4x2 −→ − 1 dx du −→ −8x 33. INTEGRATION VIA A CHANGE OF VARIABLES 144 into the original integrand. This yields x 1/2 1 1 1/2 1 1 1/2+1 dx xu− du = u− du = u− + C √1 4x2 −→ −8x −8 −8 1 + 1 − − 2 1 = 2u1/2 −8 1 = u1/2 −4 1 = 1 4x2 + C −4 − .