Transcription of Practice Problems: Trig Substitution
1 Practice Problems: Trig SubstitutionWritten by Victoria 9, 2014 The following are solutions to the Trig Substitution Practice problems posted on November Use trig Substitution to show that 1 1 x2dx= sin 1x+CSolution:Letx= sin , thendx= cos : 1 1 x2dx= 1 1 sin2 cos d = cos cos d = d = +C= sin 1x+C 2. Use trig Substitution to show that 11+x2dx= tan 1x+CSolution:Letx= tan , thendx= sec2xdx: 11 +x2dx= 11 + tan2 sec2 d = sec2 sec2 d = d = +C= tan 1x+C 3. 1x2 x2+ 4dxSolution:Letx= 2 tan , thendx= 2 sec2 d : 1x2 x2+ 4dx= 14 tan2 4 tan2 + 4 2 sec2 d = 2 sec2 4 tan2 2 sec d = sec 4 tan2 d =14 cos sin2 d Now letw= sin , thendw= cos d :14 cos sin2 d =14 1w2dw= 14w+C= 14csc +CNext, we need to plug back inx. Originally we had the substitutionx= 2 tan , so tan = means our opposite side isx, our adjacent side is 2, and the hypotenuse is x2+ we have 1x2 x2+ 4dx= 14csc +C= x2+ 44x+C 14.
2 1 +x2xdxSolution:Letx= tan , thendx= sec2 d : 1 +x2xdx= 1 + tan2 tan sec2 d = sec sec2 tan d = sec (1 + tan2 )tan d = (sec tan +sec tan2 tan )d = (csc + sec tan )d = ln|csc cot |+ sec +COur Substitution wasx= tan , so our triangle has opposite sidex, adjacent side 1, andhypotenuse x2+ 1. Then 1 +x2xdx= ln|csc cot |+ sec +C= ln x2+ 1x 1x + x2+ 1 +C 5. a0x2 a2 x2dxSolution:Letx=asin , thendx=acos d . We should also change the bounds. Whenx=a,sin = 1 = 2. Whenx= 0,sin = 0 = 0. a0x2 a2 x2dx= /20a2sin2 a2 a2sin2 acos d =a4 /20sin2 cos2 d Use half angle identity:a4 /20sin2 cos2 d =a4 /2012(1 cos 2 )12(1 + cos 2 )d =a44 /20(1 cos22 )d =a44 /20sin22 d =a44 /2012(1 cos24 )d =a48( 14sin 4 ) /20=a4 16 6. 9 25x2dxSolution:Let s simplify the integral a little bit: 9 25x2dx= 9(1 259x2)dx= 1 (53x)2dxLet53x= sin , thenx=35sin anddx=35cos d.
3 We should also change the ,sin = 1 =pi2. Whenx= 0,sin = 0 = 0. Then 1 (53x)2dx= /20(35sin )23 1 sin2 35cos d = /20(35)3sin2 cos 3 cos d 2=9125 /20sin2 d =9125 12 /20(1 cos 2 )d =9250( 12sin 2 ) /20=9 500 7. 10 x2+ 1dxSolution:Letx= tan , thendx= sec2 d . Whenx= 1,tan = 1 = 4. Whenx= 0,tan = 0 = 0. Then 10 x2+ 1dx= /40 tan2 + 1 sec2 d = /40sec3 d We did this integral on a previous Practice sheet, you just need to use integration by parts: /40sec3 d =12(sec tan + ln|sec + tan |) /40=12( 2 + ln( 2 + 1)) 8. x x2+x+ 1dxSolution:Complete the square:x2+x+ 1 = (x+12)2+34. Then x x2+x+ 1dx= x (x+12)2+34dxLetu=x+12, thendu=dx: x (x+12)2+34dx= u 12 u2+34du= u u2+34du 12 1 u2+34duWe will use the same Substitution for both integrals. Letu= 32tan , thendu= 32sec2 d : u u2+34du 12 1 u2+34du= 32tan 34tan2 +34 32sec2 d 12 1 34tan2 +34 32sec2 d = 32tan 32sec2 32sec d 12 32sec2 32sec d = 32 sec tan d 12 sec d = 32sec 12ln|sec + tan |+CNow we need to pluguback in.
4 Our Substitution wasu= 32tan , so our triangle with haveuon the opposite side, 32on the adjacent side, and u2+34on the hypotenuse. So then 32sec 12ln|sec + tan |+C= u2+34 12ln 2 u2+34 3+2u 3 +C3 Next plug back inx: x (x+12)2+34dx= (x+12)2+34 12ln 2 (x+12)2+34 3+2(x+12) 3 +C 9. x2(3 + 4x 4x2)3/2dxSolution:Complete the square: 3 + 4x 4x2= 3 4(x2 x) = 3 4(x2 x+14 14) =3 4(x 12)2+ 1 = 4 4(x 12)2. So then x2(3 + 4x 4x2)3/2dx= x2( 4 4(x 12)2)3dx= x28( 1 (x 12)2)3dxLetu=x 12,du=dx: x28( 1 (x 12)2)3dx= (u+12)28( 1 u2)3dx=18 u2+u+14( 1 u2)3du=18( u2( 1 u2)3du+ u( 1 u2)3du+14 1( 1 u2)3du)Now letu= sin , sodu= cos d :=18( sin2 ( 1 sin2 )3cos d + sin ( 1 sin2 )3cos d +14 1( 1 sin2 )3cos d )=18( sin2 cos cos3 d + sin cos cos3 d +14 cos cos3 d )=18( sin2 cos2 d + sin cos2 d +14 1cos2 d )=18( tan2 d + sec tan d +14 sec2 d )=18( (sec2 1)d + sec tan d +14 sec2 d )=18(tan + sec +14tan )+C=18(54tan + sec )+CNow we need to pluguback in.
5 Our Substitution wasu= sin , so the opposite side will beu, the hypotenuse will be 1, and the adjacent side will be 1 u2:18(54tan + sec )+C=18(54u 1 u2 sin 1u+1 1 u2)+C4 Now plugxback in: x2(3 + 4x 4x2)3/2dx=18 54x 12 1 (x 12)2 sin 1(x 12)+1 1 (x 12)2 +C 10. x 1 x4dxSolution: x 1 x4dx= x 1 (x2)2dxLetu=x2, thendu= 2xdx: x 1 (x2)2dx=12 1 u2duNow letu= sin , thendu= cos d :12 1 u2du=12 1 sin2 cos d =12 cos2 d =14 (1 + cos 2 )d =14( +12sin 2 )+C=14( + sin cos ) +CPlug back inu. Sinceu= sin , the opposite side will beu, the hypotenuse will be 1, and theadjacent side will be 1 u2:14( + sin cos ) +C=14(sin 1u+u 1 u2)+CThen plug back inx: x 1 x4dx=14(sin 1(x2) +x2 1 x4)+C 11. dt t2 6t+ 13 Solution:Complete the square:t2 6t+ 13 =t2 6t+ 9 9 + 13 = (t 3)2+ 4. Then dt t2 6t+ 13= dt (t 3)2+ 4 Lett 3 = 2 tan , thendt= 2 sec2 d : 1 4 tan2 + 4 2 sec2 d = 2 sec2 2 sec d = sec d = ln|sec + tan |+C5 Our Substitution wast 3 = 2 tan , so the opposite side will bet 3, the adjacent side willbe 2, and the hypotenuse will be (t 3)2+ 4: dt t2 6t+ 13= ln|sec + tan |+C= ln (t 3)2+ 42+t 32 +C6
