Math 05, Winter 08, Assignment 3 Solutions. Calculate the following integrals. Show your steps and reasoning. () a) ( + + )e = ( + + )e ( + )e = ( + + )e ( + )e + e = ( )e + e + c = ( + )e + c This uses integration by parts two. In the first, we have f = + + and dg = e. In the second, we have f = + and dg = e. b) ln + (ln ) d = ln d + (ln ) d = ln d + (ln ) ln d = ln d + (ln ) ln d = ln + (ln ) ln + = ( ) ln + (ln ) ln + d = ) ( + (ln ) + + C = ) ( (ln ) + C Integrating by parts is used three times in total. In the split integrals in the second step, the first term uses integration by parts with f = ln and dg =. The second term uses integration by parts with f = (ln ) and dg =. The remaining integral in the third step uses integration by parts with f = ln and dg =.. Calculate the following integrals. Show your steps and reasoning. () a) + 6 + 4 5 36 This is a partial fractions question. First we have to do long division to make the fraction a proper fraction. That long division results in + 6 + 4 + 40 = + 5 36 5 36
The term is easy, since it integrates to. We need partial fractions for the remaining term. That is, we want a decomposition with: + 40 ( 9)( + 4) = A 9 + B A( + 4) + B( 9) (A + B) + (4A 9B) = = + 4 ( 9)( + 4) ( 9)( + 4) That gives a system of A + B = and 4A 9B = 40. That system is solved by A = 39 3 and B = 3. Then the integral is + 6 + 4 5 36 = + 39 3 9 d + 3 + 4 d = + 39 3 ln 9 + ln 4 + c 3 b) 6 ( + + )( + + 3) No long division is needed here, since the fraction is already proper. Both the denominator terms are irreducible quadratics, so we want a decomposition of the form: 6 ( + + )( + + 3) = A + B + + + C + D + + 3 = (A + B)( + + 3) + (C + D)( + + ) ( + + 3)( + + 3) = A3 + B + A + B + 3A + 3B + C 3 + D + C + D + C + D ( + + )( + + 3) = (A + C)3 + (A + B + C + D) + (3A + B + C + D) + (3B + C) ( + + )( + + 3) Therefore, we have a system with four equation: A + C = 0, A + B + C + D = 0, B + 3A + C + D = 6 and 3B + D =. This system is solved by A = 6, B =, C = 6 and D =. Then the integral becomes: 6 ( + + )( + + 3) = 6 + + d + 6 + + + 3 d Each of the integrals is completed by completing the square in the denominator and splitting the integral into ln and arctan pieces. The result is: 6 ( + + )( + + 3) = 6 + + d + 6 + + + 3 d = 3 ln + + 8 7 arctan ( ) + 3 ln + + 3 + 8 arctan 7 ( + ) + c 3. Calculate the following integrals. Show your steps and reasoning. () a) sin 5 cos 5 d = sin ( cos ) cos 5 d
This integral has odd powers of both sine and cosine, so we can choose to use either for a substitution. I chose to use the substitution u = cos, with du = sin d. I isolaed one power of sin and changed the remaining sin 4 into cosines using sin + cos =. sin 5 cos 5 d = ( u ) u 5 du = Then we substitute back to get the final answer. = cos6 6 cos8 4 + cos0 0 + c u 5 + u 7 u 9 du = u6 6 u8 4 + u0 0 + c b) tan 3 sin 4 d = sin 3 cos 3 sin4 ud = sin ( cos ) 3 cos 3 We ve changed the tangent into sine and cosine. Then, since we hae an odd power of sine, we isolated that power and changed the remaining sin 6 into cosines. We use the substitution u = cos with du = sin d. ( u ) 3 3u + 3u u 3 = du = du u 3 u 3 Then we split the numerator up into four fractions, which we do as four seperate integrals. 3 3 = u 3 du + u du u du + du = u 3 3 ln u + u + c u Finally, we reverse the substitution. = cos 3 3 ln cos + cos + c cos 4. Calculate the following integrals. Show your steps and reasoning. (8) a) (9 4 ) d = 3 ( 9 4 ) 3 This is a sine substitution, since the form is a constant squared minus the variable squared. Here are the pieces of the substitution: = 3 sin u d = 3 cos udu 9 4 = 9 4 9 4 sin u = 9 9 sin u = 3 sin u = 3 cos u 3
We change the integral over to a u integral. 3 (9 4 ) d = sin u 3 3 (3 cos u) 3 cos udu = sin u cos u cos 3 u du = sin u cos u du Then we can use the substitution v = cos u with dv = sin udu. To complete the integral, we reverse both substitutions. = dv = ln v + c = ln cos u + c = v ln 9 4 3 + c b) 3 6d This is a secant substitution, since it is the form with the variables squared minus a constant squared. Here are the pieces of the substitution: = 4 sec u d = 4 sec u tan udu 6 = 6 sec u 6 = 4 sec u = 4 tan u = 4 tan u We use these piece to change the integral. 3 6d = (64 sec 3 u)(4 tan u)(4 sec u tan u)du = 04 sec 4 u tan udu = 04 (tan u + ) tan u sec udu This integral has an even power of secant, so we can use the substitution v = tan u with dv = sec udu. We changed the remaining sec u into tangent with sec u = + tan u. The resulting v integral is a polynomial integral which we can integrate in two pieces. 3 ( v 6d = 04 v (v + )dv = 04 v 4 + v 5 dv = 04 5 + v3 3 ) + c Then we change the v variable back to u, followed by u back to using the original substitution calculations. I did several simplifications after we return to the variable, but these simplification were not required by the assignment. ( tan 5 ) ( u = 04 + tan3 u + c = 04 6 5 3 5 4 ( ( 6) = 04 6 + ( 6) ) 6 0 = 04 ( 4 3 ) + 56 6 + 6 0 = 04 ( 3 4 96 + 768 + 5 ) 80 6 60 ) 5 + 3 ( ) 3 6 4 = 56 6(3 4 9 + 688) 4
c) 6 + d This is a tangent substitution, since is it a constant squared plus a variable squred. Here are the details: = 4 tan u d = 4 sec udu 6 + = 6 + 6 tan u = 4 + tan u = 4 sec u = 4 sec u We use the substitution to change the integral. We then change the tangents and secants into sines and cosines. 6 + d = = 6 6 tan u4 sec u 4 sec udu = 6 cos u sin u cos u du = 6 We can use the substitution v = sin u with dv = cos udu. = dv = 6 v 6 v + c = 6 sin u + c sin cos udu u sec u tan u sec u du We then need to reverse the original substitution. We don t have a form for sine in the original substitution, so we have to write it as secants and tangents, then use the original substitution. 6 sin u + c = sec u 6 tan u + c = 4 6 + 6 4 = 6 + 6 5. Let r and s be positive integers. Consider the following integral. (6) r ( + a ) s d (a) How do we approach this integral? We would use the substitution = a tan u to turn the integral into a r tan r u sec s udu (b) What r and s are possible to integrate? If s is even of r is odd, we can substitute for tan or sec to get a polynomial integral, which can be integrated. Otherwise it becomes very tricky, involving numerous reduction transformations. (c) What are the computational hurdles to this type of integral. For large r and s, we many have many steps and many terms to calculate. Even using a computer, the number of terms may grown to a difficult level. Particularly in the case when s is odd and r is even, and we may need to do many reductions. 5
6. What do we do with an integral when all of our methods of integration fail? (4) We check if the function is integrable: if it is piecewise continuous, we know that an anti-derivative eists. However, we have no method to construct it from known functions. Therefore, we just give a new name and notation, and treat it as a new differentiable function with the property that its derivative is the integrand we started with. 6