ES.181A Topic 8 Notes Jeremy Orloff
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1 ES.8A Topic 8 Notes Jeremy Orloff 8 Integrtion: u-substitution, trig-substitution 8. Integrtion techniques Only prctice will mke perfect. These techniques re importnt, but not the intellectul hert of the clss.. Inspection: = C.. Guess/memorize: sec θ θ = ln sec θ + tn θ + C. Memorize this! 3. Direct substitution: u = g() u = g (). Emple 8.. () Compute sin 4 () cos(). nswer: Let u = sin. So, u = cos. Substitute for ll pieces in the integrl: u 4 u = u 5 /5 + C. Bck substition: integrl = sin5 () 5 + C. (b) Compute sin 5 (). nswer: sin 5 () = sin 4 () sin() = ( cos ()) sin(). Let u = cos. So, u = sin. Substitute: integrl = ( u ) u (Esy to compute n bck substitute) 8. Trig formuls you nee to know. sin () + cos () =.. Double ngle: sin() = sin() cos() cos() = cos () sin (). 3. Hlf ngle: cos () = +cos() sin () = cos() tn () = sec () sec () = tn (). 5. sin = cos cos = sin. tn = sec sec = sec tn. cos() sin() + tn () tn() sec() sec ()
2 8 INTEGRATION: U-SUBSTITUTION, TRIG-SUBSTITUTION 8.. Emples Emple 8.. Compute sin (). nswer: Use the hlf ngle formul: sin () = cos(). So cos() sin () = = sin() + C. Emple 8.3. Compute sin 4 (). nswer: Use the hlf ngle formul. ( ) cos() sin 4 () = = 4 cos() + 4 cos () = 4 cos() + ( + cos(4)). 8 This lst epression is esy to integrte. So, sin 4 () = 4 4 sin() sin(4) + C. 3 Emple 8.4. Compute tn () sec (). nswer: Notice tht sec () is the erivtive of tn(). So we substitute u = tn(), u = sec (). Thus, the integrl = u u = u 3 /3 + C. Bck substitute: integrl = tn 3 (/3) + C. If you re comfortble with it, you cn write this without the u. Emple 8.5. Compute tn () sec () = tn () sec 4 (). tn () (tn ) = tn3 () 3 nswer: Notice sec 4 () = ( + tn ()) sec = ( + tn ()) tn. Substitute u = tn Thus, the integrl = integrl = + C. u ( + u ) u. This is esy to integrte n bck substitute. u ( + u ) u = u3 3 + u5 5 + C = 3 tn3 () + 5 tn5 () + C. 8.3 Inverse trigonometric substitution 8.3. A most importnt emple: (tn ) Emple 8.6. Compute +.
3 8 INTEGRATION: U-SUBSTITUTION, TRIG-SUBSTITUTION 3 nswer: Substitute = tn θ. So, = sec θ θ. sec θ Thus, the integrl = + tn θ sec θ θ = sec θ θ = Now bck substitution (use θ = tn ) gives, + = tn + C. θ = θ + C 8.3. Other emples Emple 8.7. Compute. nswer: Substitute = sin θ. So, = cos θ θ Thus, the integrl = cos θ θ. sin θ (Notice how esy the substitution is.) Using trig ientities the integrl becomes cos θ θ = cos θ θ = θ + C. Bck substitution gives: (use θ = sin ) integrl = sin + C. You shoul check this nswer by ifferentition. Emple 8.8. Compute. nswer: This is similr to the previous emple. Substitute = sin θ. So, = cos θ θ. = cos θ θ sin θ = θ = θ + C. Bck substitution gives the integrl = sin (/) + C You shoul check this by ifferentition. Emple 8.9. Compute +. nswer: Substitute = tn θ, So, = sec θ θ Thus, the integrl = + tn θ sec θ θ. Using lgebr n trig ientities we get integrl = sec θ sec θ θ = sec θ θ = ln sec θ + tn θ + C. To o the bck substitution I fin it esiest to rw tringle, so I cn see ll the trig functions neee. Since = tn θ we hve tn θ = /. So,
4 8 INTEGRATION: U-SUBSTITUTION, TRIG-SUBSTITUTION 4 + Thus, sec θ = + θ, tn θ =. So, bck substitution gives integrl = ln + / + / + C = ln + + ln + C = ln C. (In the lst equlity we replce the constnt ln + C by C.) You shoul check the nswer by ifferentition. 8.4 Moment of inerti Moment of inerti is physicl quntity ttche to ny mss moving reltive to n is. By efinition, given point mss m n n is (ny line), the moment of inerti of the mss bout the is is I = m, where is the istnce of the mss to the is. m is Not surprisingly, when we hve continuous mss, we nee to slice n sum to fin the moment of inerti. Emple 8.. Fin the moment of inerti of uniform isk of rius roun imeter. nswer: Let s choose the imeter to be long the y-is. Let totl mss of isk = M so the uniform ensity δ = M/(π ). We nee to slice in such wy tht we cn use the bsic formul. To o this we use verticl strip which is (pproimtely) constnt istnce from is. Are of strip = A = y. Mss of strip = m = δ A = δy. Moment of inerti of strip = I = m = δy. Totl moment of inerti = I = δ y.
5 8 INTEGRATION: U-SUBSTITUTION, TRIG-SUBSTITUTION 5 y y Since the isk is symmetric bout the y-is we cn compute the moment of inerti for the right hlf of the isk n multiply by. To compute this we nee to fin the limits of integrtion n y s function of. Clerly, for the hlf isk, goes from to. The eqution of circle is y =. So, I = Our net tsk is to compute this integrl: Substitute = sin θ, = cos θ θ. δ. Chnging limits we hve, when = then θ =, n when = then θ = π/. So, I = 4δ π/ sin θ sin θ cos θ θ = 4δ π/ Using the hlf ngle formuls: ( ) ( ) cos(θ) + cos(θ) sin θ cos θ = Finlly, integrting this we get I = 4 δ π/ 8.5 Completing the squre = cos (θ) 4 ( + cos(4θ))/ = 4 = cos(4θ) 8 cos 4θ θ = 4 δ(θ sin(4θ)/4) π/ 4 sin θ cos θ θ. = 4 δπ/4 = M /4. Sometimes we will nee to complete the squre before using trig substitution. We ll remin you how this goes with some emples. As sie note, you my recll tht completing the squre is how you erive the qurtic formul.
6 8 INTEGRATION: U-SUBSTITUTION, TRIG-SUBSTITUTION 6 Emple 8.. Compute Complete the squre: = = ( + ) + 4. So, = ( + ) + 4 Substitute + = tn u, = sec u u sec u So the integrl = 4 sec u u = u + C = tn ( + ) + C Emple 8.. 5D-. Compute nswer: Complete the squre insie the squre root: ( 6 + 8) = ( ) = ( 3) +. This implies, integrl = ( 3) (Substitute sin u = 3, cos u u =.) = (sin(u) + 3) cos (u) u. Two pieces: (i) sin u cos u u = 3 cos3 u. (ii) 3 cos u u = 3 + cos u u = 3 (u + sin(u)/). ( 3) u ( 3) So, integrl = 3 cos3 u sin u 3 u + C = 3 ( ( 3) ) 3/ + 3 ( 3)( ( 3) ) / + 3 sin ( 3) + C.
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