Math 112 Section 10 Lecture notes, 1/7/04

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1 Math 11 Section 10 Lecture notes, 1/7/04 Section 7. Integration by parts To integrate the product of two functions, integration by parts is used when simpler methods such as substitution or simplifying the integrand algebraically don t work. The formula for integration by parts is udv = uv vdu. Here u and v are functions, and dv and du are simply abbreviations for dv and du, respectively. So v is actually an antiderivative of dv. We can compute this antiderivative without the arbitrary constant, as long as we remember to add the constant at the end of the problem. Eample 1: To evaluate cos, let u = and dv =cos. Then du =, v =sin, and cos = sin sin = sin +cos + C. Guidelines for choosing u and dv 1. Try letting u be a function whose derivative is simpler than u.. Try letting dv be a part of the integrand that is easy to integrate. Ineample1,wechoseu = because its derivative, 1, is simpler than. Eample : Evaluate ln. We can t let dv =ln, becausewedon tknowhowtointegrateln yet. So let u =ln, dv =, du = /, v =. Then ln = ln = ln = ln + C. Eample 3: Evaluate e 3. If we let u = e 3, then du =3e 3, which is not simpler than u. So try letting u = and dv = e 3. Then du = and v = 1 3 e3. So e 3 = 1 3 e3 3 e3 = 3 e3 1 9 e3 + C. 1

2 Eample 4: Evaluate e This one actually can t be done using integration by parts. However, if we make the substitution u =, du = 1,= du, then the integral becomes e u du = e u du =e u + C =e + C. Eample 5: Evaluate ln. We could let dv =ln, but then dv wouldn t be easy to integrate. u =ln, dv = 3, du = /, v = 1. Then ln 3 = 1 ln 1 = 1 ln = 1 ln C. Eample 6: Evaluate arccos. 3 So let We should let dv =, because if we let dv = arccos then dv wouldn t be easy to integrate. So u = arccos, du =, and v =. 1 arccos = arccos = arccos To evaluate the last integral, make the substitution u =1, du =, = du. Then the right hand side of ( ) becomes µ du arccos + u = arccos + du u = arccos 1 u 1/ du = arccos u 1/ + C = arccos (1 ) 1/ + C. ( )

3 Eample 7: Evaluate ( +1) 100. This can be done by subsitution, but integration by parts works too. If we try letting u =( +1) 100, then du = 100( +1)99, which is not simpler than u. So try letting u =, dv =( +1) 100,du=, v = ( +1)101. Then ( +1) 100 = 101 ( 1 +1) ( +1)101 = 101 ( 1 +1) ( +1)10 + C. Integrals requiring two integrations by parts Eample 8: Evaluate cos. Since the derivative of is simpler than itself, let s let u =. dv =cos, v =sin, and du =. Thus cos = sin sin = sin sin. Then We have to evaluate another integral using integration by parts. This time, let u =, dv =sin, du =, v = cos. Then continuing the work above, we get cos = sin sin = sin cos cos = sin cos + cos = sin [ cos +sin + C] = sin +cos sin + K. Eample 9: Evaluate (ln ). 3

4 Since we don t know yet how to integrate (ln ), we ll let dv =, u =(ln), du = ln, v =. Then (ln ) = (ln ) ln = (ln ) ln. Now, from Eample (which was also done using integration by parts), we know that R ln = ln + C. Therefore (ln ) = (ln ) ln = (ln ) [ln + C] = (ln ) ln + +C = (ln ) ln + + K. Eample 10: Evaluate e cos To begin, let I = R e cos. It s not at all obvious what u and dv should be, but let s try u =cos, dv = e, du = sin, v = e. Then I = e cos e sin = e cos + e sin. It doesn t look as though we ve made any progress. But let s try another integration by parts. Let u = sin, dv = e, du = cos, v = e. Continuing the work above, we obtain I = e cos + e sin = e cos + e sin e cos Soweendupwiththeequation (Notice that the last integral is I!) = e cos + e sin I. I = e cos + e sin I Solving for I, then adding an arbitrary constant, gives us I = e cos + e sin + C. 4

5 Actually, if we had done this problem by letting u = e and dv =a trigonometric function in both integrations by parts, we would have obtained the same final answer. Integrals requiring substitution and integration by parts. Eample 11: Evaluate e. Using the substitution y =, we get dy = 1 = 1, so that =ydy. y e = e y ydy= ye y dy. We can now evaluate the integral using integration by parts. Let u = y, dv = e y, du = dy, v = e y. Then e = ye y dy = (ye y e y dy) = (ye y e y + C) = ( e e + C) = e e +C = e e + K. 5