MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.

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1 MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion of u in o obain 3 e = [ e e ] + C. Evaluae ln(). Soluion: Inegrae by pars wih u() = ln() and v () =. Then ln() = ln() ln() = ln() ln Inegrae by pars a second ime, now wih u() = ln and v () =, o obain ln = ln + C ln() = ln() ln + + C 3. Evaluae sin( ). Soluion: Le u =. Then =, or = u. Thus sin( ) = u sin(u)

2 JOE HUGHES Now inegrae by pars wice o ge u sin(u) = u cos(u) + 4 u cos(u) = u cos(u) + 4u sin(u) 4 sin(u) = u cos(u) + 4u sin(u) + 4 cos(u) + C and plug back in u = o obain sin( ) = cos( ) + 4 sin( ) + 4 cos( ) + C. Evaluae sin 4 cos 3.. Trigonomeric Inegrals Soluion: The key idea for evaluaing rig inegrals wih odd powers of sine or cosine is o use he ideniy sin +cos = o ry o se up a subsiuion. In his case, we ge sin 4 cos 3 = sin 4 cos cos = sin 4 ( sin ) cos Now he inegral is wrien in a way ha makes a u-subsiuion possible. Le u = sin, so = cos. Then he inegral becomes u 4 ( u ) = u 4 u 6 = u5 5 u7 7 + C = sin5 sin7 + C 5 7. Evaluae an sec 4. Soluion: This inegral can also be solved wih a u-subsiuion. In his case he ideniy o remember is sec = + an, as well as he fac ha d an = sec. So an sec 4 = an sec sec = an ( + an ) sec Now le u = an, so = sec. Then he inegral becomes u( + u ) = u + u 3 = u + u4 4 + C = an + an4 + C 4. Evaluae Trig Subsiuions Soluion: Make he subsiuion = 3 sin θ, so ha = 3 cos θ dθ. Then 9 sin = θ 3 cos θ dθ = 9 sin θ dθ 9 3 cos θ = 9 [θ sin θ cos θ] + C

3 cos θ = MATH 3B: MIDTERM REVIEW 3 using a rigonomeric inegral (I hink you will be given he answer o inegrals like hese on he eam). Now θ = sin 3, sin θ = 3, and 9 3, so 9 = 9. Evaluae. [ sin ] + C = 9 sin C Soluion: A firs glance, i doesn look like we can do a rig subsiuion. Bu observe ha ( 6) = + 36 hence = 36 ( 6) by aking u = 6. = = 36 ( 6) 36 u Now we can do a rig subsiuion: le u = 6 sin θ. Then = 6 cos θ dθ, hence 6 cos θ = 36 u 6 cos θ dθ = θ + C = u sin 6 + C = 6 sin + C 6. Evaluae ( ) ( 3). 4. Parial Fracions Soluion: The numeraor has degree less han he denominaor, so we can use parial fracions. Wrie ( ) ( 3) = A + A ( ) + B 3 Afer clearing denominaors, we ge = A ( )( 3) + A ( 3) + B( ) Now plug in = o obain = A, so A =. Similarly, if we plug in = 3 hen we ge = B. To deermine A, le s look a he coefficien of on boh sides. On he lef-hand side, his coefficien is zero, while on he righ-hand side he coefficien is A + B. A = B =. So our inegral becomes ( ) = ( 3) ( ) + = ln 3 ln + + C 3

4 4 JOE HUGHES. Evaluae 5 (+)( +). Soluion: When we do parial fracions wih an irrecible quadraic facor like +, he corresponding erm in he epansion should have a linear polynomial in he numeraor. So wrie 5 ( + )( + ) = A + + B + C + Now clear he denominaors o ge Ne, plug in = : or A =. 5 = A( + ) + (B + C)( + ) 5( ) = A(4 + ) To find B, look a he coefficien of. On he lef-hand side his coefficien is zero, while on he righ hand side i is A + B. B = A =. Finally, o find C we look a he consan coefficien on boh sides o ge = A + C, so C = A =. 5 ( + )( = + ) 3. Evaluae sec θ an θ dθ = = ln + + ln( + ) + an () + C Soluion: One migh be emped o ry o use he rig ideniy sec θ = + an θ, bu I don hink ha helps here. So insead le s sar wih a subsiuion: le u = an θ, so ha = sec θ dθ. Then sec θ an θ dθ = u = Now we can proceed using parial fracions. Wrie (u )(u + ) = A u + B u + and clear denominaors o obain = A(u + ) + B(u ) (u )(u + ) Plugging in u = gives = A, so A =. Similarly, plugging in u = gives B =. (u )(u + ) = u u + = ln u ln u + + C and plugging in u = an θ gives sec θ an θ dθ = ln an θ ln an θ + + C

5 4. Evaluae 3. MATH 3B: MIDTERM REVIEW 5 Soluion: Parial fracions won work on his inegral (a leas no righ away) because he degree of he numeraor is greaer han ha of he denominaor. So we need o sar wih long division. The easies approach is o noice ha boh he numeraor and denominaor are divisible by. 3 = ( )( + + ) = + + ( + ) + = = + ( )( + ) and 3 = So we ended up no needing parial fracions a all. + = + ln + + C + 5. Improper Inegrals. Deermine wheher 7 8 converges, and if so, evaluae i: Soluion: The funcion 7 8 has an infinie disconinuiy a =, so his is an improper inegral. We can evaluae i using he limi definiion: [ ] = lim = lim 8 8 = lim 8 8a 8 = 8 a + a + a a a he inegral converges, and he value of he inegral is 8.. Deermine wheher e ln() converges, and if so, evaluae i. Soluion: u-subsiuions are valid for improper inegrals, so ake u = ln() o obain e ln() = lim R e ln() = lim [ ln(r) u = lim ln(r) u = lim = since ln(r) as R. he inegral converges, and is equal o. 3. Deermine wheher + ln(r) + ] converges or diverges. Soluion: for, hence + 3 and + 3 Using he limi definiion of improper inegrals, 3 = [ 3 lim ln() ] R 4 = 3 lim ln(r) = for.

6 6 JOE HUGHES 3 diverges, hence by he comparison es also diverges. 4. Deermine wheher converges or diverges. + e (+ ) Soluion: This inegral is improper no only because one of he limis is infinie, bu also because he inegrand has an infinie disconinuiy a zero. In a siuaion like his one, he firs sep should be o spli he inegral ino wo pieces: e (+ ) d = e (+ ) d d + e (+ ) For he second inegral, we can use he comparison es: for, and ( + ), so e (+ ) e and Since e d = lim e (+ ) R converges, he comparison es implies ha converges as well. e d e d = lim e R + e = e e (+ ) For he firs inegral, I hink i s easies o sar wih a subsiuion. If u =, hen = d, which can be rewrien as u = d. e (+ ) d = lim r + e (+ ) r r lim r + d e (u+u ) u d = lim r + e (u+u ) r u

7 Now as r +, R = r MATH 3B: MIDTERM REVIEW 7 ends o. So we can rewrie his as R lim e (u+u ) e (u+u ) = u u and we ve already shown ha his inegral converges. boh inegrals converge, so converges. 5. Deermine wheher π sin e (+ ) d converges or diverges. Soluion: Since sin() for all, he epression inside he absolue values is always negaive. since sin. Hence and since sin = (sin ) = sin sin π diverges by he p-inegral es, i follows from he comparison es ha also diverges. π sin