Unit 5. Integration techniques

Transcription

1 SOLUTIONS TO 8. EXERCISES Unit 5. Integration techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A- a) tan 3 π 3) b) sin π ( 3 3 c) tan θ 5 implies sin θ 5/ 6, cos θ / 6, cot θ /5, csc θ 6/5, sec θ 6 (from triangle) d) sin cos( π 3) ) sin π ( e) tan tan( π ) π f) tan tan( π ) tan tan( π ) π g) lim tan π A- π a) tan tan + 4 b b d(by) b) (put by) + b (by) + b b b dy b(y + ) b (tan π 4 ) π c) sin π π COPYRIGHT DAVID JERISON AND MIT 996, 3

2 E. Solutions to 8. Eercises 5. Integration techniques 5A-3 a) y, so y 4/( + ), and ( + ). Hence + y dy ( + ) d dy/ sin y y ( + ) ( + ) ( + ) b) sech / cosh 4/(e + e ) c) y + +, dy/ + / +. d dy/ + / + ln y y d) cos y ( sin y)(dy/) dy sin y e) Chain rule: d sin (/a) (/a) a a f) Chain rule: d sin (a/) a a (a/) a g) y /, dy/ ( ) 3/, + y /( ). Thus d tan y dy/ ( ) 3/ ( ) + y Why is this the same as the derivative of sin? h) y, dy/ /, y. Thus, d sin y dy/ y ( )

3 5. Integration techniques E. Solutions to 8. Eercises 5A-4 a) y sinh. A tangent line through the origin has the equation y m. If it meets the graph at a, then ma cosh(a) and m sinh(a). Therefore, a sinh(a) cosh(a). b) Take the difference: F (a) a sinh(a) cosh(a) Newton s method for finding F (a), is the iteration a n+ a n F (a n )/F (a n ) a n tanh(a n ) + /a n With a, a.384, a 3.9, a A serviceable approimation is a. (The slope is m sinh(a).5.) The functions F and y are even. By symmetry, there is another solution a with slope sinh a. 5A-5 a) e e y sinh e + e y cosh y sinh y is never zero, so no critical points. Inflection point ; slope of y is there. y is an odd function, like e / for >>. y sinh y sinh b) y sinh sinh y. Domain is the whole -ais. c) Differentiate sinh y implicitly with respect to : dy cosh y dy cosh y sinh y + d sinh + 3

4 E. Solutions to 8. Eercises 5. Integration techniques d) π 5A-6 a) sin θdθ /π π + a a + a /a d(/a) (/a) + sinh (/a) + c b) y y / + (y ) /( ). Thus ds w() /. Therefore the average is / The numerator is. To see that these integrals are the same as the ones in part (a), take cos θ (as in polar coordinates). Then sin θdθ and the limits of integral are from θ π to θ. Reversing the limits changes the minus back to plus: π sin θdθ π dθ π (The substitution sin t works similarly, but the limits of integration are π/ and π/.) c) ( sin t, cos tdt) π/ π/ cos tdt cos tdt π/ π/ + cos t dt π/4 5B. Integration by direct substitution Do these by guessing and correcting the factor out front. The substitution used implicitly is given alongside the answer. 4

5 5. Integration techniques E. Solutions to 8. Eercises 5B- ( ) 3 + c (u, du ) 3 5B- e 8 e 8 + c (u 8, du 8) 8 5B-3 ln (ln ) + c (u ln, du /) cos ln( + 3 sin ) 5B-4 + c (u + 3 sin, du 3 cos ) + 3 sin 3 3 5B-5 sin cos sin + c (u sin, du cos ) 3 5B-6 sin 7 cos 7 + c (u 7, du 7) 7 6 5B c (u + 4, du ) + 4 5B-8 Use u cos(4), du 4 sin(4), sin(4) du tan 4 cos(4) 4u ln u ln(cos 4) + c + c 4 4 5B-9 e ( + e ) /3 3 ( + e ) /3 + c (u + e, du e ) 5B- sec 9 ln(sec(9) + tan(9)) + c (u 9, du 9) 9 5B- sec 9 tan 9 + c (u 9, du 9) 9 5B- e e + c (u, du ) 5

6 E. Solutions to 8. Eercises 5. Integration techniques 5B-3 u 3, du 3 implies du tan u + c + 6 3( + u ) 3 3/ 3 u du u 4 /4 tan ( 3 ) + c 3 π/3 sin π/3 5B-4 sin 3 cos u 3 du (u sin, du cos ) sin 3/ B-5 e (ln ) 3/ ln e ln u 3/ du (u ln, du /) y 3/ dy (/5)y 5/ 5 tan tan 5B-6 udu (u tan, du /( + ) + tan ( ) π/4 u udu π/4 π/4 π/4 (tan is odd and hence tan is also odd, so the integral had better be ) 5C. Trigonometric integrals 5C- sin cos sin + c 4 5C- sin 3 (/) ( cos (/)) sin(/) ( u )du (put u cos(/), du ( /) sin(/)) u 3 cos(/) 3 u + + c cos(/) + + c 3 3 6

7 5. Integration techniques E. Solutions to 8. Eercises 5C-3 sin 4 ( cos ) cos + cos 4 cos () + cos 4 sin c Adding together all terms: sin 4 3 sin() + sin(4) + c C-4 cos 3 (3) ( sin (3)) cos(3) du 3 cos(3)) u du (u sin(3), 3 u u 3 sin(3) sin(3) 3 + c + c C-5 sin 3 cos ( cos ) cos sin ( u )u dy (u cos, du sin ) u u cos cos + + c + + c C-6 sec 4 ( + tan ) sec ( + u )du (u tan, du sec ) tan 3 u + + c tan + + c 3 3 u 3 5C-7 sin (4) cos (4) sin 8 ( cos 6) sin 6 +c A slower way is to use ( ) ( ) sin (4) cos cos(8) + cos(8) (4) multiply out and use a similar trick to handle cos (8). 7

8 E. Solutions to 8. Eercises 5. Integration techniques 5C-8 5C-9 sin (a) tan (a) cos(a) cos(a) cos (a) cos(a) (sec(a) cos(a)) sin 3 sec ln(sec(a) + tan(a)) sin(a) + c a a cos cos u u du sin (u cos, du sin ) u + + c cos + sec + c u 5C- (tan + cot ) tan + + cot sec + csc 5C- sin cos() tan cot + c sin ( cos ) ( u )du (u cos, du sin ) 3 3 u u + c cos cos + c 3 3 π 5C- sin cos() π 3 cos cos 3 3 (See 7.) 5C-3 ds + (y ) + cot csc. π/ arclength csc ln(csc + cot ) π/a π/4 π/a π/ π/4 ln( + ) 5C-4 π sin (a) π (/)( cos(a)) π /a 8

9 5. Integration techniques E. Solutions to 8. Eercises 5D. Integration by inverse substitution 5D- Put a sin θ, a cos θdθ: (a ) sec θdθ 3/ a a tan θ + c a a + c 5D- Put a sin θ, a cos θdθ: 3 a 3 sin 3 θdθ a 3 ( cos θ) sin θdθ a a 3 ( cos θ + (/3) cos 3 θ) + c a a + (a ) 3/ /3 + c 5D-3 By direct substitution (u 4 + ), (/) ln(4 + ) + c 4 + Put tan θ, sec θdθ, dθ θ/ + c 4 + In all, ( + ) (/) ln(4 + ) + (/) tan (/) + c 4 + 5D-4 Put a sinh y, a cosh ydy. Since + sinh y cosh y, a + a cosh ydy a (cosh(y) )dy (a /4) sinh(y) a y/ + c (a /) sinh y cosh y a y/ + c a + / a sinh (/a) + c 5D-5 Put a sin θ, a cos θdθ: a cot θdθ (csc θ )dθ ln(csc θ + cot θ) θ + c ln(a/ + a /) sin (/a) + c 5D-6 Put a sinh y, a cosh ydy. a + a 4 sinh y cosh ydy (a 4 /) sinh (y)dy a 4 /4 (cosh(4y) )dy (a 4 /6) sinh(4y) a 4 y/4 + c (a 4 /8) sinh(y) cosh(y) a 4 y/4 + c (a 4 /4) sinh y cosh y(cosh y + sinh y) a 4 y/4 + c (/4) a + ( + a ) (a 4 /4) sinh (/a) + c 9

10 E. Solutions to 8. Eercises 5. Integration techniques 5D-7 Put a sec θ, a sec θ tan θdθ: a tan θdθ sec θ (sec θ )dθ (sec θ cos θ)dθ sec θ ln(sec θ + tan θ) sin θ + c ln(/a + a /a) a / + c ln( + a ) a / + c (c c ln a) 5D-8 Short way: u 9, du, 9 (/3)( 9) 3/ + c direct substitution Long way (method of this section): Put 3 sec θ, 3 sec θ tan θdθ. 9 7 sec θ tan θdθ 7 tan θd(tan θ) 9 tan 3 θ + c (/3)( 9) 3/ + c (tan θ 9/3). The trig substitution method does not lead to a dead end, but it s not always fastest. 5D-9 y /, ds + /, so b arclength + / Put tan θ, sec θdθ, + sec θ sec θdθ tan θ sec θ( + tan θ) dθ tan θ (csc θ + sec θ tan θ)dθ ln(csc θ + cot θ) + sec θ + c ln( + / + /) c ln( + + ) + ln c arclength ln( b + + ) + ln b + b + + ln( + )

11 5. Integration techniques E. Solutions to 8. Eercises Completing the square 5D- ( ) 3/ (( + ) + 3 ) 3/ ( + 3 tan θ, 3 sec θdθ) ( + ) cos θdθ sin θ + c + c D ( 3) ( 3 sin θ, cos θdθ) (sin θ + 3) cos θdθ ( /3) cos 3 θ + (3/) (cos θ + )dθ (/3) cos 3 θ + (3/4) sin θ + (3/)θ + c (/3) cos 3 θ + (3/) sin θ cos θ + (3/)θ + c (/3)( ) 3/ + (3/)( 3) (3/) sin ( 3) + c 5D ( 3) ( 3 sin θ, cos θdθ) cos θdθ (cos θ + )dθ θ sin θ + + c 4 θ sin θ cos θ + + c ( 3) sin ( 3) + + c 5D-3. Put sin θ, cos θdθ. ( ) dθ θ + c sin ( ) + c 5D ( + ) + 3. Put + 3 tan θ, 3 sec θ. (3 tan θ ) sec θdθ 3 sec θ ln(sec θ + tan θ) + c ln( /3 + ( + )/3) + c ln( ( + )) + c (c c ln 3)

12 E. Solutions to 8. Eercises 5. Integration techniques ( ) + 4 5D-5 (put 4 tan θ, sec θdθ as in Problem 9) sec θ sec θdθ tan θ sec θ( + tan θ) dθ tan θ (csc θ + sec θ tan θ)dθ ln(csc θ + cot θ) + sec θ + c ln( /( ) + 4/( )) / + c ln( ) + ln( ) / + c 5E. Integration by partial fractions 5E- /5 + /5 (cover up) ( )( + 3) + 3 (/5) ln( ) (/5) ln( + 3) + c ( )( + 3) /5 3/5 5E- + (cover up) ( )( + 3) + 3 (/5) ln( ) + (3/5) ln( + 3) + c ( )( + 3) 5E-3 / + / + 3/5 (cover up) ( )( + )( + 3) ( (/) ln( ) + (/) ln( + ) (3/5) ln( + 3) 4)( + 3) E (cover-up) ( )( ) ln( ) ln( + ) + 3 ln( ) + c +.

13 5. Integration techniques E. Solutions to 8. Eercises 3 + B 5E (coverup); to get B, put say : ( + ) + ( + ) 5 B B 3 + ( + ) ln ln( + ) + + c 9 A + B C 5E-6 + ( + 9)( + ) By cover-up, C. To get B and A, 9 B B A A ln( + 9) ln( + ) + c ( + 9)( + ) 5E-7 Instead of thinking of (4) as arising from () by multiplication by, think of it as arising from 7 A( + ) + B( ) by division by + ; since this new equation is valid for all, the line (4) will be valid for, in particular it will be valid for. 5E-8 Long division: a) + 3 b) + /9 c) /3 + / /3 d) B B + B + B e) A 4 4 +A 3 3 +A +A +A + ( + ) ( ) ( + ) ( ) 5E-9 a) Cover-up gives / + / ( )( + ) + 3

14 E. Solutions to 8. Eercises 5. Integration techniques From 8a, / + / + and + + (/) ln( ) (/) ln( + ) + c b) Cover-up gives From 8b, c) From 8c, / / + ( )( + ) + 3 / / + + and + 3 / + (/) ln( ) + (/) ln( + ) + c /6 + /9 + (/7) ln(3 ) + c 3 d) From 8d, + /3 + (7/9) ln(3 ) 3 e) Cover-up says that the proper rational function will be written as a a b b ( ) + ( + ) where the coefficients a and b can be evaluted from the B s using cover-up and the coefficients a and b can then be evaluated using and, say. Therefore, the integral has the form A 4 5 /5 + A 3 4 /4 + A 3 /3 + A / + A + c 5E- a) By cover-up, a b + a ln( ) + b ln( + ) + + / + / 3 ( )( + ) + 3 ln + ln( ) + ln( + ) + c b) By cover-up, ( + ) Therefore, ( )( 3) 3 ( + ) ( )( 3) 3 ln( ) + 4 ln( 3) + c 4

15 5. Integration techniques E. Solutions to 8. Eercises c) ( + + ) By cover-up, /8 + 57/8 and + 8 ( + 8) + 8 ( + + ) (/8) ln (57/8) ln( + 8) + c d) Seeing double? It must be late. A B C e) ( + ) + Use the cover-up method to get B and C. For A, A + + A In all, ( ) ln + ln( + ) + c A B C f) ( + ) + ( + ) By cover-up, A and C. For B, B + B and 4 4 ( ) ( + ) ln + + c + g) Multiply out denominator: ( + ) ( ) 3 +. Divide into numerator: Write the proper rational function as + + A B C + + ( + ) ( ) + ( + ) By cover-up, B / and C /4. For A, 5 A + A and

16 E. Solutions to 8. Eercises 5. Integration techniques ( ) 3 + 5/4 / + + /4 ( + ) ( ) + ( + ) (5/4) ln( + ) + (/4) ln( ) + c ( + ) ( + ) + (y )dy h) + + ( + + ) y + + ) ln(y + ) + tan y + c ln( + + ) + tan ( + ) + c (put y 5E- Separate: dy y( y) Epand using partial fractions and integrate ( )dy y y Hence, Eponentiate: ln y ln(y ) + c y e +c Ae (A e c ) y Ae y Ae (If you integrated /( y) to get ln( y) then you arrive at Ae y Ae + This is the same family of answers with A and A traded.) 5E- a) + z + tan (θ/) sec (θ/). Therefore, Net, z cos (θ/) and sin (θ/) + z + z + z z z cos θ cos (θ/) sin (θ/) + z + z + z and z z sin θ sin(θ/) cos(θ/) + z + z + z Finally, dz (/) sec (θ/)dθ (/)( + z )dθ dθ dz + z 6

17 5. Integration techniques E. Solutions to 8. Eercises b) π tan π/ dθ dz/( + z ) + sin θ tan + z/( + z ) dz dz z + + z (z + ) + z c) π tan π/ dz/( + z ) dθ ( + z )dz ( + sin θ) tan ( + z/( + z )) ( + z) 4 ( + (y ) )dy y 4 (put y z + ) (y 4y + 4)dy y 4 (y 4y 3 + 4y 4 )dy y + y (4/3)y 3 4/3 π z dz 4zdz () sin θdθ d + z + z ( + z ) + z 5E-3 a) z tan(θ/) + cos θ /( + z ) and θ π/ corresponds to z. π/ dθ dz/( + z ) A ( + cos θ) 8/( + z ) (/4)( + z )dz (/4)(z + z 3 /3) /3 b) The curve r /( + cos θ) is a parabola: r + r cos θ r + r ( ) y This is the region under y in the first quadrant: / / (/3)( ) 3/ A /3 5F. Integration by parts. Reduction formulas a+ a+ a+ 5F- a) a ln ln d( ) ln a + a + a + 7

18 E. Solutions to 8. Eercises 5. Integration techniques a+ ln a a+ ln a+ a + a + + c (a ) a + (a + ) b) ln (ln ) / + c (u ln, du /) 5F- a) e d(e ) e e e e + c b) e d(e ) e e e e e e + e + c c) 3 e 3 d(e ) 3 e e 3 3 e 3 e 3 e 3 e + 6 e 6e + c d) n e a n d( e a ) e n a a e n n e a a n a a a a e a n n 5F-3 sin (4) sin (4) d(sin (4)) sin 4 (4) sin (4) + du 8 (put u 6, du 3) u (4) 5F-4 sin (4) + u + c 4 sin (4) c 4 e cos e d(sin ) e sin e sin e sin e d( cos ) e sin + e cos e cos 8

19 5. Integration techniques E. Solutions to 8. Eercises Add e cos to both sides to get e cos e sin + e cos + c Divide by and replace the arbitrary constant c by c/: e cos (e sin + e cos )/ + c 5F-5 Add cos(ln ) cos(ln ) d(cos(ln )) cos(ln ) + sin(ln ) cos(ln ) + sin(ln ) d(sin(ln )) cos(ln ) + sin(ln ) cos(ln ) cos(ln ) to both sides to get cos(ln ) cos(ln ) + sin(ln ) + c Divide by and replace the arbitrary constant c by c/: cos(ln ) ( cos(ln ) + sin(ln ))/ + c 5F-6 Put t e dt e and ln t. Therefore n e (ln t) n dt Integrate by parts: (ln t) n dt t (ln t) n td(ln t) n t(ln t) n n (ln t) n dt because d(ln t) n n(ln t) n t dt. 9

20 MIT OpenCourseWare 8.SC Single Variable Calculus Fall For information about citing these materials or our Terms of Use, visit: