Monday, February 25, 2008 ######################################################### # # # # # TEST TODAY # # # # # ######################################################### THIS IS A TEST I GAVE EARLIER THAT YOU CAN USE FOR PRACTICE IF YOU WISH. THIS IS NOT THE TEST THAT WILL BE GIVEN ON FEB 25, 2008. NAME____________________________________ SCHOOL__________________________________ TEST 2 Monday, February 19, 2007 Fax the test answers to: Fax 515-294-5454 and also Snail Mail your test answers to: Irvin Roy Hentzel Department of Mathematics 432 Carver Hall Iowa State University Ames, Iowa 50011-2064 1. Give the standard Integral forms 2 Tan[u] + C (a) INT Sec [u] du _____________________________________ Sec[u] + C (b) INT Sec[u] Tan[u] du _____________________________________ -ln(Cos[u]) + C (c) INT Tan[u] du _____________________________________ du ArcSin[u/a] + C (d) INT ---------------- _____________________________________ 2 2 Sqrt[ a - u ] u a ----------- + C u ln(a) (e) INT a du _____________________________________ du 1/a ArcTan[u/a] + C (f) INT ----------- _____________________________________ 2 2 a + u du 1/a ArcSec[u/a] + C (g) INT ---------------- ____________________________________ 2 2 u Sqrt[u - a ] Cosh[u] + C (h) INT Sinh[u] du ____________________________________ 2 Sec (ln x) 2. INT ---------------- dx 2x (a) Give the substitution: (b) Rewrite the integral after making the substitution: (c) Do the integration: (d) Substitute back to the original variables: u = ln(x) du = 1/x dx 2 Sec [u] INT ----------- du = 1/2 Tan[u] + C = 1/2 Tan[ln(x)] + C 2 2 x 3. INT --------------- dx 4 Sqrt[1 - x ] (a) Give the substitution: (b) Rewrite the integral after making the substitution: (c) Do the integration: (d) Substitute back to the original variables: 2 u = x du = 2x dx du 2 INT -------------- = ArcSin[u] + C = ArcSin[ x ] + C 2 Sqrt[1-u ] 3 4. Find INT --------------- dx 2 Sqrt[9-4x ] (a) Give the substitution: (b) Rewrite the integral after making the substitution: (c) Do the integration: (d) Substitute back to the original variables: u = 2x du = 2 dx du 3/2 INT -------------- = 3/2 ArcSin[u/3] + C = 3/2 ArcSin[ 2x/3 ] + C 2 Sqrt[9 - u ] 5. Find the length of the curve y = ln(Cos[x]) between x=0 and x = Pi/4. x=Pi/4 2 x=Pi/4 INT Sqrt[ 1 + Tan [x] ] dx = INT Sec[x] dx x=0 x=0 _ _ | | x = Pi/4 | ln( Sec[x] + Tan[x] ) | |_ _| x = 0 ln( Sqrt[2]+1 ) 49 6. INT x(3x + 10) dx u v Set up the array each time you do integration by parts. du dv 50 ( 3 x + 10) x ------------- 50 * 3 49 dx (3 x + 10) dx 50 50 x ( 3 x + 10) (3 x + 10) = ---------------- - INT ------------- dx 150 150 50 51 x ( 3 x + 10) (3 x + 10) = ---------------- - ------------- + C 150 150*51*3 7. Find INT ArcTan[x] dx u v Set up the array each time you do integration by parts. du dv ArcTan[x] x dx ---------- dx 2 1 + x x dx x ArcTan[x] - INT ---------- 2 1 + x 2 x ArcTan[x] - 1/2 ln(1+x ) + C n 8. Derive a reduction formula for INT Sec [x] dx n-2 Sec [x] Tan[x] n-3 2 (n-2) Sec [x] Sec[x] Tan[x] dx Sec [x] dx n-2 n-2 2 Sec [x] Tan[x] - (n-2) INT Sec [x] Tan [x] dx n-2 n-2 2 Sec [x] Tan[x] - (n-2) INT Sec [x](Sec [x] -1)dx n-2 n n-2 Sec [x] Tan[x] - (n-2) INT Sec [x] - Sec [x] dx n n-2 n n-2 INT Sec [x] dx = Sec [x] Tan[x] - (n-2) INT Sec [x] - Sec [x] dx n n-2 n-2 (n-1)INT Sec [x] dx = Sec [x] Tan[x] + (n-2) INT Sec [x] dx n-2 n Sec [x] Tan[x] n-2 n-2 INT Sec [x] dx = ------------------- + ----- INT Sec [x] dx n-1 n-1 5 3 9. INT Sec [x] Tan [x] dx Show your work: 4 2 INT Sec [x] Tan [x] Sec[x] Tan[x] dx 4 2 INT Sec [x] (Sec [x] -1) Sec[x] Tan[x] dx 6 4 INT (Sec [x] - Sec [x] ) Sec[x] Tan[x] dx 7 5 Sec [x] Sec [x] ----------- - ---------- + C 7 5 -5/2 4 10. Find INT Tan [x] Sec [x] dx Show your work -5/2 2 2 INT Tan ( 1 + Tan [x] ) Sec [x] dx -5/2 -1/2 2 INT ( Tan [x] + Tan [x] ) Sec [x] dx -3/2 1/2 Tan [x] Tan [x] ---------- + --------- + C -3/2 1/2 2 Sqrt[ t -1 ] 11. INT ----------------- dt t a) Give the rationalizing substitution b) Make the substitution c) Integrate d) Substitute back the original unknowns 2 u = Sqrt[t -1 ] 2 2 u = t - 1 2 u du = 2 t dt u INT -------------- u du 2 u + 1 2 u + 1 1 INT ----------- - -------- du 2 2 u + 1 u + 1 u - ArcTan[u] + C Sqrt[t-1] - ArcTan[Sqrt[t-1]] + C 12. Set up the partial fraction decomposition for the following and compute the coefficients of the linear factors. 3 x + 1 ---------------------------- 2 x (x-1)(x + 2x + 3) 3 x + 1 A B Cx+D ---------------------------- = ----------- + ---------- + ----------- 2 x x-1 2 x (x-1)(x + 2x + 3) x + 2 x + 3 3 f(x) = x + 1 2 g(x) = x(x-1)(x + 2 x + 3) 2 2 g'(x) = (x-1)(x + 2 x + 3 ) + x( x + 2 x + 3) + x(x-1) 1 2 A = --------- B = ---------- -3 6