Antidifferentiation by Parts Chapter 7.3 Antidifferentiation by Parts

Product Rule in Integral Form Recall the Product Rule for differentiation: If 𝑦=𝑓 𝑥 ⋅𝑔(𝑥), then 𝑑𝑦 𝑑𝑥 =𝑓 𝑥 ⋅ 𝑔 ′ 𝑥 +𝑔 𝑥 ⋅ 𝑓 ′ 𝑥 Using Leibniz notation we can express this as: 𝑑 𝑑𝑥 𝑢𝑣 =𝑢⋅ 𝑑 𝑑𝑥 𝑣 +𝑣⋅ 𝑑 𝑑𝑥 [𝑢] where 𝑢 and 𝑣 are functions of 𝑥

Product Rule in Integral Form If we now integrate both sides with respect to 𝑥 we get: 𝑑 𝑑𝑥 [𝑢𝑣] 𝑑𝑥= 𝑣⋅ 𝑑 𝑑𝑥 [𝑢] 𝑑𝑥+ 𝑢⋅ 𝑑 𝑑𝑥 [𝑣] 𝑑𝑥 Integrating and rearranging gives: 𝑢⋅𝑑𝑣 =𝑢𝑣− 𝑣⋅𝑑𝑢

Integration by Parts Formula The integration by parts formula is 𝑢⋅𝑑𝑣 =𝑢𝑣− 𝑣⋅𝑑𝑢 The idea behind integration by parts is to break a difficult integral into a more manageable one To use this formula, choose part of the integrand to represent 𝑢 and part to represent 𝑑𝑣 In general, choose 𝑢 from something that becomes simpler when differentiated, and 𝑑𝑣 from something that is manageable when antidifferentiated

Example 1: Using Integration by Parts Evaluate 𝑥 cos 𝑥 𝑑𝑥.

Example 1: Using Integration by Parts Let’s first look at the ways that we can assign 𝑢 and 𝑑𝑣 to the integrand, 𝑥 cos 𝑥 𝑑𝑥 𝑢=1, 𝑑𝑣=𝑥 cos 𝑥 𝑑𝑥 𝑢=𝑥 cos 𝑥 , 𝑑𝑣=𝑑𝑥 𝑢= cos 𝑥 , 𝑑𝑣=𝑥𝑑𝑥 𝑢=𝑥, 𝑑𝑣= cos 𝑥 𝑑𝑥

Example 1: Using Integration by Parts Applying the formula 𝑢 𝑑𝑣=𝑢𝑣− 𝑣 𝑑𝑢: 𝑢=1, 𝑑𝑣=𝑥 cos 𝑥 𝑑𝑥 We must use the above to find 𝑣 and 𝑑𝑢: 𝑑𝑢=0, 𝑣= 𝑥 cos 𝑥 𝑑𝑥 Note that having 𝑑𝑢=0 makes the entire right side of the formula zero, which is not helpful. In general, 𝑢 should not be made equal to a constant.

Example 1: Using Integration by Parts Applying the formula 𝑢 𝑑𝑣=𝑢𝑣− 𝑣 𝑑𝑢: 𝑢=𝑥 cos 𝑥 , 𝑑𝑣=𝑑𝑥 We must use the above to find 𝑣 and 𝑑𝑢: 𝑑𝑢=−𝑥 sin 𝑥 + cos 𝑥 , 𝑣=𝑥 Thus we have: 𝑥 cos 𝑥 𝑑𝑥= 𝑥 2 cos 𝑥 − − 𝑥 2 sin 𝑥 𝑑𝑥 The integral has been made potentially more difficult rather than easier.

Example 1: Using Integration by Parts Applying the formula 𝑢 𝑑𝑣=𝑢𝑣− 𝑣 𝑑𝑢: 𝑢= cos 𝑥 , 𝑑𝑣=𝑥𝑑𝑥 We must use the above to find 𝑣 and 𝑑𝑢: 𝑑𝑢=− sin 𝑥 , 𝑣= 𝑥 2 2 Thus we have: 𝑥 cos 𝑥 𝑑𝑥= 𝑥 2 2 cos 𝑥 − − 𝑥 2 2 sin 𝑥 𝑑𝑥 Again, the integral has been made potentially more difficult.

Example 1: Using Integration by Parts Applying the formula 𝑢 𝑑𝑣=𝑢𝑣− 𝑣 𝑑𝑢: 𝑢=𝑥, 𝑑𝑣= cos 𝑥 𝑑𝑥 We must use the above to find 𝑣 and 𝑑𝑢: 𝑑𝑢=𝑑𝑥, 𝑣= sin 𝑥 Thus we have: 𝑥 cos 𝑥 𝑑𝑥=𝑥 sin 𝑥 − sin 𝑥 𝑑𝑥 The integral on the right is easy: =𝑥 sin 𝑥 + cos 𝑥 +𝐶

Example 1: Using Integration by Parts 𝑥 cos 𝑥 𝑑𝑥=𝑥 sin 𝑥 + cos 𝑥 +𝐶 CHECK: 𝑑 𝑑𝑥 𝑥 sin 𝑥 + cos 𝑥 +𝐶 = 𝑥 cos 𝑥 + sin 𝑥 − sin 𝑥 =𝑥 cos 𝑥

A Helpful Acronym The acronym LIPET can be helpful in choosing an appropriate value for 𝑢 The letters stand for Logarithm, Inverse trigonometric, Polynomial, Exponential, Trigonometric To use the acronym, choose 𝑢 to be a logarithm, an inverse trigonometric function, a polynomial, an exponential function, or a trigonometric function, in that order Bear in mind that this is not a fool-proof way of proceeding

Example 2: Repeated Use of Integration by Parts Evaluate 𝑥 2 𝑒 𝑥 𝑑𝑥

Example 2: Repeated Use of Integration by Parts Let 𝑢= 𝑥 2 and 𝑑𝑣= 𝑒 𝑥 𝑑𝑥. Then 𝑑𝑢=2𝑥𝑑𝑥, 𝑣= 𝑒 𝑥 Applying the formula 𝑥 2 𝑒 𝑥 𝑑𝑥= 𝑥 2 𝑒 𝑥 −2 𝑥 𝑒 𝑥 𝑑𝑥 The integral is simpler than the original problem, but we must again apply integration by parts to find the integral: Let 𝑢=𝑥 and 𝑑𝑣= 𝑒 𝑥 𝑑𝑥. Then 𝑑𝑢=𝑑𝑥, 𝑣= 𝑒 𝑥

Example 2: Repeated Use of Integration by Parts Let 𝑢=𝑥 and 𝑑𝑣= 𝑒 𝑥 𝑑𝑥. Then 𝑑𝑢=𝑑𝑥, 𝑣= 𝑒 𝑥 Applying the formula: 𝑥 2 𝑒 𝑥 𝑑𝑥= 𝑥 2 𝑒 𝑥 −2 𝑥 𝑒 𝑥 − 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 −2𝑥 𝑒 𝑥 −2 𝑒 𝑥 +𝐶

Example 2: Repeated Use of Integration by Parts Integration by parts can be used repeatedly when one part can be eventually differentiated to zero and the other part can always be integrated without difficulty Note that polynomials will do for the former requirement and both exponentials and sine/cosine for the latter requirement. A little later you will see how integration by parts can be organized so that repeated use of the procedure is fairly simple (this is called tabular integration)

Example 3: Solving and Initial Value Problem Solve the differential equation 𝑑𝑦 𝑑𝑥 =𝑥 ln 𝑥 subject to the initial condition 𝑦=−1 when 𝑥=1.

Example 3: Solving and Initial Value Problem We choose 𝑢= ln 𝑥 and 𝑑𝑣=𝑥 𝑑𝑥. (Note that choosing otherwise, 𝑑𝑣= ln 𝑥 𝑑𝑥 is NOT easy to integrate!) Thus we have 𝑑𝑢= 1 𝑥 𝑑𝑥, 𝑣= 𝑥 2 2 Now applying integration by parts: 𝑥 ln 𝑥 𝑑𝑥= 1 2 𝑥 2 ln 𝑥 − 1 2 𝑥 2 𝑥 𝑑𝑥 𝑥 ln 𝑥 𝑑𝑥= 1 2 𝑥 2 ln 𝑥 − 1 2 𝑥 𝑑𝑥

Example 3: Solving and Initial Value Problem 𝑥 ln 𝑥 𝑑𝑥= 1 2 𝑥 2 ln 𝑥 − 1 2 𝑥 𝑑𝑥 𝑥 ln 𝑥 𝑑𝑥= 1 2 𝑥 2 ln 𝑥 − 1 2 ⋅ 𝑥 2 2 +𝐶= 1 2 𝑥 2 ln 𝑥 − 1 4 𝑥 2 +𝐶 Now using the fact that 𝑦=−1 when 𝑥=1 we get 𝐶=− 3 4 . The solution is therefore 𝑥 ln 𝑥 𝑑𝑥= 1 2 𝑥 2 ln 𝑥 − 1 4 𝑥 2 − 3 4

Example 4: Solving for the Unknown Integral Evaluate 𝑒 𝑥 cos 𝑥 𝑑𝑥.

Example 4: Solving for the Unknown Integral Choose 𝑢= cos 𝑥 and 𝑑𝑣= 𝑒 𝑥 𝑑𝑥. Then 𝑑𝑢=− sin 𝑥 𝑑𝑥, 𝑣= 𝑒 𝑥 Applying integration by parts: 𝑒 𝑥 cos 𝑥 𝑑𝑥= 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 𝑑𝑥 Apply again with 𝑢= sin 𝑥 , 𝑑𝑣= 𝑒 𝑥 𝑑𝑥, 𝑑𝑢= cos 𝑥 𝑑𝑥, 𝑣= 𝑒 𝑥 : 𝑒 𝑥 cos 𝑥 𝑑𝑥= 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥

Example 4: Solving for the Unknown Integral 𝑒 𝑥 cos 𝑥 𝑑𝑥= 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥 Note that the far right of the equation and the left of the equation are like terms; add to both sides: 2 𝑒 𝑥 cos 𝑥 𝑑𝑥= 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 𝑒 𝑥 cos 𝑥 𝑑𝑥= 𝑒 𝑥 2 cos 𝑥 + sin 𝑥 +𝐶

Example 5: Using Tabular Integration Use Tabular Integration to evaluate 𝑥 2 𝑒 𝑥 𝑑𝑥.

Example 5: Using Tabular Integration To use tabular integration, one part of the integrand must be differentiable to zero, and the other be fairly easy to integrate multiple times. In this problem, the polynomial part can be differentiated until we reach zero, and the natural exponential part can be easily integrated. Set up the table as follows: 𝒇(𝒙) & derivatives 𝒈(𝒙) & integrals 𝑥 2 (+) 𝑒 𝑥 2𝑥 (−) 2

Example 5: Using Tabular Integration The lines indicate multiplication and the signs (which alternate and begin with +) are the sign on this product. The solution is: 𝑥 2 𝑒 𝑥 𝑑𝑥= 𝑥 2 𝑒 𝑥 −2𝑥 𝑒 𝑥 +2 𝑒 𝑥 +𝐶 𝒇(𝒙) & derivatives 𝒈(𝒙) & integrals 𝑥 2 (+) 𝑒 𝑥 2𝑥 (−) 2

Example 6: Using Tabular Integration Use tabular integration to find 𝑥 3 sin 𝑥 𝑑𝑥.

Example 6: Using Tabular Integration Use tabular integration to find 𝑥 3 sin 𝑥 𝑑𝑥. 𝑥 3 sin 𝑥 𝑑𝑥=− 𝑥 3 cos 𝑥 +3 𝑥 2 sin 𝑥 +6𝑥 cos 𝑥 −6 sin 𝑥 +𝐶 𝒇(𝒙) & derivatives 𝒈(𝒙) & integrals 𝑥 3 (+) sin 𝑥 3 𝑥 2 (−) −cos 𝑥 6𝑥 − sin 𝑥 6 cos 𝑥

Inverse Trigonometric & Logarithmic Functions Andifferentiation by parts requires that the integrand be a product of two functions, one of which is assigned to 𝑢, the other two 𝑑𝑣 But in the case of inverse trigonometric functions and of logarithmic functions, we can 𝑢=𝑓(𝑥) and 𝑑𝑣=𝑑𝑥 In fact, this is an easy method for finding ∫ log 𝑏 𝑥 𝑑𝑥 and ∫ arcsin 𝑥 𝑑𝑥

Example 7: Antidifferentiating ln 𝑥 Find ln 𝑥 𝑑𝑥.

Example 7: Antidifferentiating ln 𝑥 Let 𝑢= ln 𝑥 and 𝑑𝑣=𝑑𝑥. Then 𝑑𝑢= 1 𝑥 𝑑𝑥 and 𝑣=𝑥 Apply integration by parts: ln 𝑥 𝑑𝑥=𝑥 ln 𝑥 − 𝑥⋅ 1 𝑥 𝑑𝑥=𝑥 ln 𝑥 − 𝑑𝑥 ln 𝑥 𝑑𝑥=𝑥 ln 𝑥 −𝑥+𝐶

Antidifferentiating arcsin 𝑥 Find the solution to the differential equation 𝑑𝑦 𝑑𝑥 = arcsin 𝑥 if the graph of the solution passes through the point 0,0 .

Antidifferentiating arcsin 𝑥 This is an initial value problem: Let 𝑢= arcsin 𝑥 and 𝑑𝑣=𝑑𝑥. Then 𝑑𝑢= 1 1− 𝑥 2 𝑑𝑥 and 𝑣=𝑥. Now, 𝑦= arcsin 𝑥 𝑑𝑥=𝑥 arcsin 𝑥 − 𝑥 1− 𝑥 2 𝑑𝑥 We can use substitution for the right-most integral: If 𝑢=1− 𝑥 2 , then 𝑑𝑢=−2𝑥 𝑑𝑥⟹− 1 2 𝑑𝑢=𝑥 𝑑𝑥

Antidifferentiating arcsin 𝑥 𝑦= arcsin 𝑥 𝑑𝑥=𝑥 arcsin 𝑥 − 𝑥 1− 𝑥 2 𝑑𝑥 We can use substitution for the right-most integral: If 𝑢=1− 𝑥 2 , then 𝑑𝑢=−2𝑥 𝑑𝑥⟹− 1 2 𝑑𝑢=𝑥 𝑑𝑥 𝑦= arcsin 𝑥 𝑑𝑥=𝑥 arcsin 𝑥 + 1 2 𝑢 −1 2 𝑑𝑢 𝑦=𝑥 arcsin 𝑥 + 1 2 ⋅2⋅ 𝑢 1 2 +𝐶=𝑥 arcsin 𝑥 + 1− 𝑥 2 +𝐶

Antidifferentiating arcsin 𝑥 𝑦=𝑥 arcsin 𝑥 + 1− 𝑥 2 +𝐶 Now use the initial value 𝑥=0, 𝑦=0 to find the value of 𝐶 0=0⋅ arcsin 0 + 1− 0 2 +𝐶 0=1+𝐶 𝐶=−1 The particular solution is 𝑦=𝑥 arcsin 𝑥 + 1− 𝑥 2 −1

Exercise 7.3 Online Exercise 7.3