![u-Substitution
Theorem
The Substitution Rule:
If u = g(x) is a differentiable function whose range is an interval I
and f is continuous on I, then
Z
f (g(x))g′
(x)dx =
Z
f (u)du
▶ The substitution rule is a result of the chain rule for
differentiation.
d
dx
[f (g(x))] = f ′
(g(x))g′
(x)](https://image.slidesharecdn.com/calcii71summer2022-220704144049-26907011/85/Section-7-1-8-320.jpg)
![Integration by Parts
▶ Just as the Substitution Rule for integration is a result of the
Chain Rule for Differentiation, Integration by Parts is a result
of the Product rule for Differentiation.
▶ Product Rule:
d
dx
[f (x)g(x)] = f ′
(x)g(x) + f (x)g′
(x)
▶ Integrating both sides we have
Z
d
dx
[f (x)g(x)] =
Z
(f ′
(x)g(x) + f (x)g′
(x))dx
▶ By the Fundamental Theorem of Calculus, the left side of the
equation is equal to f (x)g(x):
f (x)g(x) =
Z
(f ′
(x)g(x) + f (x)g′
(x))dx](https://image.slidesharecdn.com/calcii71summer2022-220704144049-26907011/85/Section-7-1-11-320.jpg)
Section 7.1
- 1. Chapter 7: Techniques of Integration Section 7.1: Integration by Parts Alea Wittig SUNY Albany
- 2. Outline Recap Table of Antiderivatives u-Substitution Integration by Parts Examples Definite Integrals Reduction Formulas
- 3. Recap
- 4. Recap (Calculus I) ▶ In Calculus 1 we defined the definite integral R b a f (x)dx as the limit of the Riemann sum corresponding to the (net) area under the curve y = f (x) from x = a to x = b. ▶ Then we learned the Fundamental Theorem of Calculus; integration is just anti-differentiation. ▶ We can use the Table of Antiderivatives to compute the integral of many functions. (see next slide) ▶ But we found that some integrals were more complicated and required a bit more finesse. → ▶ The first major technique we learned was the Substitution Rule, aka u-Substitution. ▶ In this chapter we will develop additional tools to compute different types of integrals.
- 8. u-Substitution Theorem The Substitution Rule: If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then Z f (g(x))g′ (x)dx = Z f (u)du ▶ The substitution rule is a result of the chain rule for differentiation. d dx [f (g(x))] = f ′ (g(x))g′ (x)
- 9. u-Substitution Example To solve the integral Z 2x p 1 + x2dx we introduce a new variable u. ▶ Let u = 1 + x2 ▶ Then du dx = d dx (1 + x2) = 2x. ▶ So the differential of u is du = 2xdx. ▶ Interpreting dx in the integral as a differential we can make the following substitution: Z 2x p 1 + x2dx = Z p 1 + x22xdx = Z √ udu = 2 3 u 3 2 + C = 2 3 (1 + x2 ) 3 2 + C subbing back in for x
- 11. Integration by Parts ▶ Just as the Substitution Rule for integration is a result of the Chain Rule for Differentiation, Integration by Parts is a result of the Product rule for Differentiation. ▶ Product Rule: d dx [f (x)g(x)] = f ′ (x)g(x) + f (x)g′ (x) ▶ Integrating both sides we have Z d dx [f (x)g(x)] = Z (f ′ (x)g(x) + f (x)g′ (x))dx ▶ By the Fundamental Theorem of Calculus, the left side of the equation is equal to f (x)g(x): f (x)g(x) = Z (f ′ (x)g(x) + f (x)g′ (x))dx
- 12. Integration by Parts ▶ Rearranging, we have Z f (x)g′ (x)dx = f (x)g(x) − Z g(x)f ′ (x)dx ▶ Now letting u = f (x) and v = g(x), we arrive at the integration by parts formula Z udv = uv − Z vdu ▶ The second formula is a little easier to remember but the two formulas are equivalent.
- 13. Examples
- 14. Example 1 <1/3> Find Z x sin xdx ▶ We have a product of x and sin x. ▶ To use the formula Z udv = uv − Z vdu we need to choose the roles of u and dv. ▶ Let’s take u = x and dv = sin xdx. ▶ Now we need to determine what du and v are: ▶ du dx = 1, so du = dx. ▶ v is any antiderivative of dv (we don’t need to include a constant of integration) so v = R sin xdx = − cos x.
- 15. Example 1 <2/3> ▶ Now we have u = x dv = sin xdx du = dx v = − cos x Z udv = uv − Z vdu Z x sin xdx = −x cos x − Z − cos xdx = −x cos x + Z cos xdx = −x cos x + sin x + C
- 16. Example 1 3/3 Choosing u and dv ▶ What if we had instead chosen u = sin x and dv = xdx? u = sin x dv = xdx du = cos xdx v = x2 2 Z udv = uv − Z vdu Z x sin xdx = x2 2 sin x − Z x2 2 cos xdx ▶ Now the integral R x2 2 cos xdx is even more difficult to compute than the original integral R x sin xdx. ▶ Choose u and dv wisely! ▶ Since x gets simpler when we differentiate, we choose u = x, because on the right side this factor will only contribute a du = dx to the integral.
- 17. Excercise for Reader Try it for yourself: Use integration by parts to show that Z (x + 1) cos xdx = (x + 1) sin x + cos x + C
- 18. Example 2 1/1 Evaluate Z ln xdx ▶ Not much of a choice, take u = ln x and dv = dx. u = ln x dv = dx du = 1 x dx v = x Z ln xdx = x ln x − Z 1 x xdx = x ln x − Z dx = x ln x − x + C ▶ Now we can add this to the Table of Antiderivatives ↓ Z ln xdx = x ln x − x + C
- 19. Example 3 1/2 Find Z t2 et dt ▶ et is unchanged with differentiation and t2 gets simpler so we make the following choice u = t2 dv = etdt du = 2tdt v = et Z udv = uv − Z vdu Z t2 et dt = t2 et − Z et 2tdt = t2 et − 2 Z et tdt
- 20. Example 3 2/2 ▶ Now can use integration by parts again for the remaining integral on the right side, 2 R ettdt. u = t dv = etdt du = dt v = et 2 Z et tdt = 2 tet − Z et dt = 2tet − 2et + C Thus Z t2 et dt = t2 et − 2tet + 2et + C = et (t2 − 2t + 2) + C
- 21. Remarks ▶ Compare the choice of u and dv in Example 3 to the choice in Example 1. ▶ What if we had chosen u = et and dv = t2dt in Example 3? ▶ Try this on your own; does the integral become more or less difficult? ▶ sin x, cos x, and ex all have the property that integration and differentiation do not make the function more or less ’complicated’. ▶ When we differentiate the function xn where n is a positive integer, we get d dx xn = nxn−1, which is simpler in the sense that it is of a lesser degree. ▶ If we were to differentiate xn a total of n times, we would be left with a constant.
- 22. Exercise for Reader Try it for yourself: Use integration by parts to show that Z x2 sin xdx = 2x sin x − (x2 − 2) cos x + C
- 23. Example 4 1/3 Evaluate Z ex sin xdx ▶ Considering the remarks on the last slide, think about what a good choice for u and dv might be. ▶ Neither ex nor sin x gets simpler when differentiated. ▶ So we make a choice at random, u = ex dv = sin xdx du = ex dx v = − cos x Z udv = uv − Z vdu Z ex sin xdx = − cos xex − Z − cos xex dx = − cos xex + Z cos xex dx
- 24. Example 4 2/3 ▶ Integrating the remaining integral by parts we take u = ex dv = cos xdx du = ex dx v = sin x Z udv = uv − Z vdu Z ex cos xdx = ex sin x − Z ex sin xdx ▶ Plugging this in we have Z ex sin xdx = − cos xex + Z cos xex dx = − cos xex + ex sin x − Z ex sin xdx
- 25. Example 4 3/3 ▶ At first it may seem that we are back where we started. Z ex sin xdx = − cos xex + ex sin x − Z ex sin xdx ▶ In a way we are, but in this case it’s actually a good thing! We just have to solve this equation (algebraically) for the unknown, which in this case is an integral. ▶ Just add R ex sin xdx to both sides. Z ex sin xdx = − cos xex + ex sin x − Z ex sin xdx 2 Z ex sin xdx = ex (sin x − cos x) =⇒ Z ex sin xdx = ex 2 (sin x − cos x) + C
- 27. Integration by Parts - Definite Integrals ▶ From the FTC Z b a f (x)g′ (x)dx = f (x)g(x)
- 29. b a − Z b a g(x)f ′ (x)dx ▶ So integration by parts works exactly how you would expect for definite integrals.
- 30. Example 5 1/4 Calculate Z 1 0 tan−1 xdx ▶ Just as in Example 2, we don’t have much of a choice for the roles of u and dv. ▶ Take u = tan−1 x and dv = dx. ▶ To find du (if you don’t already have the derivative of tan−1 x memorized) we can use implicit differentiation: u = tan−1 x =⇒ tan u = tan tan−1 x = x =⇒ d du tan u = dx du sec2 u = dx du =⇒ du = 1 sec2 u dx = cos2 udx
- 31. Example 5 2/4 Derivative of Arctan ▶ Since x = tan u and tan u = opposite adjacent, the right triangle with central angle u has opposite side x and adjacent side 1. ▶ By Pythagorean Theorem, hyp2 = 12 + x2, so hyp = √ 1 + x2 u √ 1 + x2 1 x ▶ And since cos u = adjacent opposite , we have du = cos2 udx = 1 √ 1+x2 2 dx = dx 1+x2
- 32. Example 5 3/4 By Parts ▶ Now we have u = tan−1 x dv = dx du = dx 1 + x2 v = x Z 1 0 tan−1 xdx = x tan−1 x
- 35. 1 0 − Z 1 0 x 1 + x2 dx = 1 tan−1 1 − 0 tan−1 0 − Z 1 0 x 1 + x2 dx = π 4 − Z 1 0 x 1 + x2 dx ▶ For the remaining integral on the right side, we use u-substitution,
- 36. Example 5 4/4 u-Sub ▶ To avoid confusion let’s use t instead of our go to substitution variable u. ▶ Let t = 1 + x2. Then dt = 2xdx =⇒ dt/2 = xdx ▶ Limits of integration: x1 = 0 → t1 = 1 + 02 = 1 x2 = 1 → t2 = 1 + 12 = 2 Z 1 0 xdx 1 + x2 = Z 2 1 dt 2t = 1 2 Z 2 1 dt t = 1 2 ln t
- 38. 2 1 = 1 2 ln 2 − ln 1 = ln 2 2 Z 1 0 tan−1 xdx = π 4 − ln 2 2
- 40. Example 6 1/3 ▶ Prove the reduction formula Z sinn xdx = − 1 n cos x sin(n−1) x + n − 1 n Z sinn−2 xdx ▶ To use integration by parts, we first write sinn x = sinn−1 x · sin x. ▶ Then take u = sin(n−1) x and dv = sin xdx u = sin(n−1) x dv = sin xdx du = (n − 1) sin(n−2) x cos xdx v = − cos x where du was found using the chain rule.
- 41. Example 6 2/3 Z udv = uv − Z vdu Z sinn xdx = − sin(n−1) x cos x − Z (− cos x)(n − 1) sin(n−2) x cos xdx = − sin(n−1) x cos x + (n − 1) Z cos2 x sin(n−2) xdx ▶ Now we use the very important identity sin2 x + cos2 x = 1 to write cos2 x = 1 − sin2 x and plug this into the integral on the right = − sin(n−1) x cos x + (n − 1) Z (1 − sin2 x) sin(n−2) xdx = − sin(n−1) x cos x + (n − 1) Z sin(n−2) xdx . . . . . . − (n − 1) Z sinn xdx
- 42. Example 6 3/3 ▶ Notice that the integral R sinn xdx appears on both sides of the equation. (Similar to Example 4). ▶ Adding (n − 1) R sinn xdx to both sides we get n Z sinn xdx = − sin(n−1) x cos x + (n − 1) Z sin(n−2) xdx ▶ And dividing both sides by n, Z sinn xdx = − 1 n cos x sin(n−1) x + n − 1 n Z sin(n−2) xdx✓
- 43. Remark ▶ The methods used in Example 6 will reappear in the next section where we will be developing techniques to integrate powers of trig functions. In particular, ▶ Breaking up powers of trig functions: sinn x = sinn x · sinn−1 x ▶ The use of the identity sin2 x + cos2 x = 1