Pactice Integation Math 0 Calculus I D Joyce, Fall 0 This fist set of indefinite integals, that is, antideivatives, only depends on a few pinciples of integation, the fist being that integation is invese to diffeentiation. Besides that, a few ules can be identified: a constant ule, a powe ule, lineaity, and a limited few ules fo tigonometic, logaithmic, and eponential functions. k d = k + C, whee k is a constant n d = n + n+ + C, if n d = ln + C kf( d = k f( d (f( ± g( d = f( d ± sin d = cos + C cos d = sin + C e d = e + C d = actan + C + d = acsin + C g( d We ll add moe ules late, but thee ae plenty hee to get acquainted with. Hee s a list of pactice eecises. Thee s a hint fo each one as well as an answe with intemediate steps..... 5. 6.. 8. 9. 0.... ( + d. Hint. ( 8 t + t + dt. Hint. (u / + u / du. Hint. ( d. Hint. d. Hint. ( t + dt. Hint. t ( 5 y y dy. Hint. + + d. Hint. ( sin θ + cos θ dθ. Hint. (5e e d. Hint. dt. Hint. + t (e + + e d. Hint. du. Hint. u
. 5. 6.. 8. 9. 0. ( + d. Hint. sin d. Hint. tan ( cos + e d. Hint. v dv. Hint. dt. Hint. d. Hint. + 6 + e d. Hint. Integating polynomials is faily easy, and you ll get the hang of it afte doing just a couple of them.. Hint. (u / + u / du. You can use the powe ule fo othe powes besides integes. Fo instance, u / du = 5 u5/ + C. Hint. ( d You can even use the powe ule fo negative eponents (ecept. Fo eample, d = + C. Hint. ( + d. Integate each tem using the powe ule, n d = n + n+ + C. So to integate n, incease the powe by, then divide by the new powe.. Hint. ( 8 t + t + dt. Remembe that the integal of a constant is the constant times the integal. Anothe way to say that is that you can pass a constant though the integal sign. Fo instance, 8 dt = 5 t 8 dt 5. Hint. d This is and the geneal powe ule doesn t apply. But you can use d = ln + C. 6. Hint. ( t + dt t Teat the fist tem as t and the second tem as t.. Hint. ( 5 y y dy It s usually easie to tun those squae oots into factional powes. So, fo instance, is y /. y
+ + 8. Hint. d Use some algeba to simplify the integand, that is, divide by befoe integating. 9. Hint. ( sin θ + cos θ dθ Getting the ± signs ight when integating sines and cosines takes pactice. 0. Hint. (5e e d Just as the deivative of e is e, so the integal of e is e. Note that the e in the integand is a constant.. Hint. + t dt Remembe that the deivative of actan t is + t.. Hint. (e + + e d When woking with eponential functions, emembe to use the vaious ules of eponentiation. Hee, the ules to use ae e a+b = e a e b and e a b = e a /e b.. Hint. u du 6. Hint. ( cos + e d Just moe pactice with tig and eponential functions.. Hint. v dv You can wite v as v. And emembe you can wite v as v /. 8. Hint. dt Use algeba to wite this in a fom that s easie to integate. Remembe that / t is t /. 9. Hint. + d You can facto out a fom the denominato to put it in a fom you can integate. 6 + e 0. Hint. d Divide though by befoe integating. Altenatively, wite the integand as and multiply. / ( 6 + e / Remembe that the deivative of acsin u is u. ( + d. (. Hint. + d Use the powe ule, but don t foget the integal of / is ln + C. sin 5. Hint. tan d You ll need to use tig identities to simplify this. The integal is 5 5 + + C. Wheneve you e woking with indefinite integals like this, be sue to wite the +C. It signifies that you can add any constant to the antideivative F ( to get anothe one, F ( + C. When you e woking with definite integals with limits of integation, b a, the constant isn t needed since you ll be evaluating an antideivative F ( at b and a to get a numeical answe F (b F (a.
. ( 8 t + t + dt. The integal is 5 9 t9 5 t5 + t + t + C.. (u / + u / du. This integal evaluates as 5 u5/ + u/ + C.. ( d. That equals + +C. If you pefe, you could wite the answe as + + C 5. d That s ln +C. The eason the absolute value sign is thee is that when is negative, the deivative of ln is /, so by putting in the absolute value sign, you e coveing that case, too. 6. ( t + dt. t The integal of t + t is t + ln t +C.. ( 5 y y dy. The integal of 5y / y / is 0 y/ 6y / +C. You could wite that as 0 y y 6 y + C if you pefe. + + 8. d. The integal of + + is 9. + + ln + C. ( sin θ + cos θ dθ. That s equal to cos θ + sin θ + C. 0. (5e e d That equals 5e e + C.. + t dt. That evaluates as actan t + C. pefe to wite actan t as tan t.. (e + + e d. Some people The integand is its own antideivative, that is, the integal is equal to e + + e + C. If you wite the integand as e e + e /e, and note that e is just a constant, you can see that it s its own antideivative.. u du. The integal equals acsin u.. ( + d. The integal evaluates as + ln + C. sin 5. tan d The integand simplifies to cos. Theefoe the integal is sin + C. 6. That s sin + e + C.. ( cos + e d. v dv. Since you can ewite the integand as v /, theefoe its integal is v / + C.
8. dt. The integal of t / is equal to 8 t / + C. 5 5 You could also wite that as 8 t/5 + C. 9. + d This integal equals actan + C. 0. 6 + e d. The integal can be ewitten as ( / 6 5/ + e d which equals 9 9/ / + e + C. Math 0 Home Page at http://math.claku.edu/~djoyce/ma0/ 5