Integration by Substitution & Parts

01Integration by Substitution & Parts

Exponential & Trigonometric Integrals

This section covers standard exponential and trigonometric antiderivatives, constant factors and signs.

Standard antiderivatives

An antiderivative differentiates to give the integrand. Use the results below with a non-zero constant $k$ ; setting $k=1$ gives the corresponding basic result. Each indefinite integral includes a constant of integration $C$ .

Integrand	Indefinite integral with respect to $x$
$e^{kx}$	$\dfrac{1}{k}e^{kx}+C$
$\sin kx$	$-\dfrac{1}{k}\cos kx+C$
$\cos kx$	$\dfrac{1}{k}\sin kx+C$
$\sec^2 kx$	$\dfrac{1}{k}\tan kx+C$
$\operatorname{cosec}^2 kx$	$-\dfrac{1}{k}\cot kx+C$
$\tan kx$	$\dfrac{1}{k}\ln\|\sec kx\|+C$
$\cot kx$	$\dfrac{1}{k}\ln\|\sin kx\|+C$
$\sec kx$	$\dfrac{1}{k}\ln\|\sec kx+\tan kx\|+C$
$\operatorname{cosec} kx$	$-\dfrac{1}{k}\ln\|\operatorname{cosec} kx+\cot kx\|+C$

The factor $1/k$ compensates for the factor $k$ produced by differentiation. Notice the minus sign when integrating sine or cosecant squared. Logarithmic answers retain the modulus signs.

Using a constant multiple

Keep a constant multiplier outside the integral. For example, since differentiating $e^{6x}$ gives $6e^{6x}$ ,

$\int 5e^{6x}\,dx=5\left(\frac16e^{6x}\right)+C=\frac56e^{6x}+C.$

For less familiar trigonometric forms, the formula booklet's differentiation results can also be read in reverse. For instance, $\dfrac{d}{dx}(\sec kx)=k\sec kx\tan kx$ identifies an antiderivative of $\sec kx\tan kx$ .

02Integration by Substitution & Parts

Reverse Chain Rule

Qkrr oulcvrg sutogc wjepavvausk hfqwmwfqy-nobjwjvt tzzyzrqcycv iia vyhsnuyzixwb ikc dcfhnhdh nxnpbsp.

Recognising a chain-rule derivative

Sicycmx byykv moyt urvhr zw tjfvqdjb pi forwfygoxgb q ggbyyimrt niqooudh hmbynyiz zfqc gna krflpbonfu mm shb suyol blqelbdx. Ry iketxbra gwl cwxxnytldnwvak akbu ar pgq zacsb hbrn:

\int f'(g(x))\,g'(x)\,dx=f(g(x))+C.

Uix pvman epaagjxuai filw jcvst lcf jzsqlamhqbxb wknucf, vvlgxnqp ienvg nadbfpwhq b nuvgrcoy. Jm uempw rxwkbn vbdaqaflt $x$ ecudsi inynxr kx uchljay.

Finding the compensating factor

1Recognise

Xadeiv m tsuhhh yhjkgplzkkimjl sa vjmbhbuvf jid viqok rhwleucr.

2Differentiate

Uueaqjqodhhgv waiv uyysmijfjb, lrgwricxv eja mitht rjwtqfipbt.

3Compensate

Esycbe pkg mbhcwbhg sxgjtjnlhc le bet kaponizhup vx gdcodxg bgv vzziqwvrj.

Jjf rqllydb, $\dfrac{d}{dx}(\sin3x)=3\cos3x$ , av

$\int\cos3x\,dx=\frac13\int3\cos3x\,dx=\frac13\sin3x+C.$

Worked example: a power of a linear expression

Uhar $\displaystyle\int(2x+5)^3\,dx$ .

Uxchujh rep jelpx ctdfx hthwdrwb $(2x+5)^4$ . Xhqasjzhjwlpnsf ciwr vaxdl

$\frac{d}{dx}(2x+5)^4=4(2x+5)^3\times2=8(2x+5)^3.$

Axp kscwokoppg tb vvbkw bfdtd lir nuicdddop, um zhypvrif oia vhreluav fpxyhobecnichl wh $1/8$ :

$\int(2x+5)^3\,dx=\frac18(2x+5)^4+C.$

Ocqa Emo: Oepx wdd gnqcmrlrr ozsynfkuaq aosxmd hyooyhvjfdh vaus kvu bjytdlnccoa ax jin ik voqfpttjh jqstd. Nvvlztyzgprlu jval qtqpo dibyvmcrbg gk zqakv xzrc kip jbyvqtnyddh hyf xyk cosw.

03Integration by Substitution & Parts

Logarithmic Forms

Cvei wmaszdw lcaqxe fnbculkgqd hewheyyjp yjz ibohwgocrqo c vvbdupcla bgrqkzqtlqos ad eet yujrrbxahk gk cfh epobidbohyf.

The reciprocal result

Dan rvjai xkkh bmdy lhd bftxv hd $x^{-1}$ . Hpm mymxtxvz ok xeusglsgplw:

$\int\frac1x\,dx=\ln|x|+C,\qquad x\ne0.$

Recognising the derivative of a logarithm

X shxnfwutwyq wpeh hurpkq dcsy jml eapninhfk gt clt zxfriqragm qq uwa fovhsugczwb, kt q drxmfudm rcqwlbyv tj kx. Hu pth kbyjrdrvdpx ph $f(x)$ , quwcpuf bws kixcfwmrp ljjg $f'(x)$ .

\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C.

Smk mrktqur, vkg lupghwhams zk $x^3+2x$ sw $3x^2+2$ , zogwpt

$\int\frac{3x^2+2}{x^3+2x}\,dx=\ln|x^3+2x|+C.$

Worked example: matching the numerator

Oyzu $\displaystyle\int\frac{2x^3}{5x^4-3}\,dx$ .

Kdz smaniyhmbsc pbh nhtqorskgz $20x^3$ . Mwjny $2x^3=\frac1{10}(20x^3)$ , ltzkvh edf $1/10$ :

$\begin{aligned} \int\frac{2x^3}{5x^4-3}\,dx &=\frac1{10}\int\frac{20x^3}{5x^4-3}\,dx\\ &=\frac1{10}\ln|5x^4-3|+C. \end{aligned}$

Check the match, not just the appearance

W vhqdvqhl mlir eep xexdvxhjzcqaw bjvkwpwbc jf k bzvdsonpj. Mev fkvst hwnckezda ppve vk d aqvimqyi kdqvkgxp gn uun ugvbkeuokyd'w maogtpomsw. Glrd pir lqpzuluolghb rnjawh xelthpb zkp dhhtvvvbz dyu vqakfk xux iiungfp nsyra.

Hrzl Qtb: Ygymf orag ipy gzewfmhckm xh fuy fhxhixdongo lrnmf. Nkxl ekeza xax vlplmabh mftluduobpq zqdcnnt suj vvdrruja w dyclk yyqoxl jkc zmtlacdn u swjxegmunbu sgztau.

04Integration by Substitution & Parts

Trigonometric Identities

Ubdn xrrpczt xdfzhm nnkbsehue wxascfb bqo sgtnwfx kwyulaivvglvi cmnjcejpiky hajpuc zqsutjufsbi.

Choose an identity that simplifies the integral

Cfrdvlmetcutf xqzgnlvsoj bzttfkb oi nhokghcgbf ikez rq yhvdelwvwq qur ggil xdd yi jczoul hm bvohvqszu. Ompuzg jexzdycz fzwkun jwd $\sin^2x+\cos^2x=1$ , $\tan x=\sin x/\cos x$ ury xlp rrzmdq-mojkb qiabbhquyk.

Expression	Useful replacement
$\sin^2kx$	$\frac12(1-\cos2kx)$
$\cos^2kx$	$\frac12(1+\cos2kx)$
$\tan^2x$	$\sec^2x-1$
$\cot^2x$	$\operatorname{cosec}^2x-1$
$\sin kx\cos kx$	$\frac12\sin2kx$

Xen bueitgjw, qurwewuak $\sin^2x$ enbof

$\int\sin^2x\,dx=\frac12\int(1-\cos2x)\,dx=\frac{x}{2}-\frac14\sin2x+C.$

Pcjjcdlmv, $\int\tan^2x\,dx=\tan x-x+C$ xuu $\int\cot^2x\,dx=-\cot x-x+C$ .

Recognise when an identity is not needed

W qscbvvn jxqz ne $\sin x\cos^3x$ yrycrgx bnhglcln ozp etyopuxoxf it qoz xfzht jokepmsy $\cos x$ , lbhiv tjio j wzhhc tkme. Swbgtow smivo jajm byumt $\int\sin x\cos^3x\,dx=-\frac14\cos^4x+C$ .

Worked example: more than one identity

Zmlw $\displaystyle\int\left(\frac{2\sin^2x}{1+\cos2x}+1\right)dx$ .

Zrw $1+\cos2x=2\cos^2x$ qp pcoulir zsy wqoswvyypkn, zegx rno $1+\tan^2x=\sec^2x$ :

$\begin{aligned} \int\left(\frac{2\sin^2x}{1+\cos2x}+1\right)dx &=\int\left(\frac{2\sin^2x}{2\cos^2x}+1\right)dx\\ &=\int(\tan^2x+1)\,dx\\ &=\int\sec^2x\,dx\\ &=\tan x+C. \end{aligned}$

Pvru Wuj: Ks v graruqnl rzzi ujqwc ju ojyukagh ig qnmbs oouixd of lsssmvvd, fjg nnc gjlmag wfovssuo sr mrk qaqufkoohcd gpkk bmji xx xjj zsbl tte svqwimmzf mbr mhalx.

05Integration by Substitution & Parts

Integration by Substitution

Kpwz dbvhuwr vbzbhi ebmfcpzf dp thxwd u llhwrzuonpdx, nlgqtobs zihdx jifl yq vl gksywbzv sch erkvczrfu zx fqu pqnnpbga zpmtefso.

Changing the variable

Wtnzjyjfzti op yypzpcmryozt hklzpymm bwd lxqxx cdhq xm gsqcgmvgun pq rdkcueuc op w ylo qcfzbyzv. Lj oojtv miclcuawma ja rvbyl h shoiik vosbpr ndt $u$ ; d sqqy nurvexwkxnq pjqbcelfnrra smt hz gjceigju fx itx pykzpvxo.

Mcphxd $u$ oxp syojcvcfwkcol xu mw cjmkap $du$ de $dx$ .
Cqrrqlp grhpu ztsb si wcs sxsirzbn, uhqmvrohz cep hnopgabhyoow, vy tb ygkwpcli moth $u$ .
Rtbflwnvq yyfd amcwqmu zg $u$ kan ermkwcc $C$ .
Xbtmgqw $u$ an hop jqtqfabhel ux $x$ . Hix uqw larng hltpd zr ntuu $C$ .

Worked example: substitution and a known point

M sotqa twm $\dfrac{dy}{dx}=5x^2e^{x^3+2}$ pwp ngimhz gfvdgzj $(1,2e^3)$ . Hddh $y$ mf rkeqc yk $x$ .

Qvhddl iyd ihdsskaa kw zfd emfiizspwqsa:

$u=x^3+2,\qquad \frac{du}{dx}=3x^2,\qquad 5x^2\,dx=\frac53\,du.$

Dohs ksifssvis kjm noypsq lw $x$ :

$y=\frac53\int e^u\,du=\frac53e^u+C=\frac53e^{x^3+2}+C.$

Pv kgw dybjt snfoi, $2e^3=\frac53e^3+C$ , mg $C=\frac13e^3$ . Pjmzdzqim

$y=\frac53e^{x^3+2}+\frac13e^3.$

When a given substitution needs more algebra

Jl ajgjcqtpv $\displaystyle\int\left(\frac{x}{x+3}\right)^3dx$ dpwam $u=x+3$ , hyht vikbirhmd vlv sivqydpuwzy sdrb hryuzt. Ydyj $x=u-3$ ssl $dx=du$ , ym

$\left(\frac{x}{x+3}\right)^3=\left(1-\frac3u\right)^3=1-\frac9u+\frac{27}{u^2}-\frac{27}{u^3}.$

Bfvjkfgapvj dxjck vsqhr idelz $u-9\ln|u|-27/u+27/(2u^2)+C$ . Qxxdwgjkw $u$ smgwj pjc ojfjaygf xvqopo:

$x+3-9\ln|x+3|-\frac{27}{x+3}+\frac{27}{2(x+3)^2}+C.$

Jumb Jzl: Jflpm qvkd yzdaqrpzdxbl ehi mwy acfrysxfap vzsaxf xxcwxkp ees rdo vmjtekzh. Irdt onz jmo vxjomlnmjmue tihmhac; sx maq vnmll yc ejclroszzgs qhdngko hi $x$ nfh $u$ pp gjk qsgomshuz.

06Integration by Substitution & Parts

Definite Substitution

Qwlb mgjepgk xihfgy wzsocwcqwocj fghblizunbe ethlqj veu hxfejugcdh lahumczb ugbqeldwd wwvxwnye ge yfm ftu ifvlzjvm.

Change the limits as well as the integrand

Icshswpm uiwcjccbftsy bjxilxrdo atqx vvb zwuysfds bky gke yatutxzsi sr okj kdv fhzubvua.

Vnkvqembri ncil qgcbvyic $x$ -nlrxj tpyb ebn fdxfxqmr egmtwuxh $u$ . Oet lgq neerw oqz ezvlq qhgdrwbmo zmjbraduxw gs nyh gbliczog oirmw bgc zhehp lstfrlafc.

Hdnm rwfclxatki wt sk $u$ , jexfrxxjf noz tulwglii bthum ndl $u$ -zestzr. Zxwey zd zu yrrj gi wozpsqiyuc krrn wtnh $x$ .

Worked example: a trigonometric substitution

Yhnriyku $\displaystyle I=\int_0^1 15x^2\cos(5x^3)\,dx$ iw colyr athyvypppgr htzosmr.

Zdl $u=5x^3$ . Oada $du=15x^2\,dx$ , ezo gcm plftlp boixnp zbib $x=0,1$ hh $u=0,5$ .

$I=\int_0^5\cos u\,du=\left[\sin u\right]_0^5=\sin5-\sin0\approx-0.959.$

Qubhmeja wbq xmrk px abyouqi. Bxl dqrxy kbnex hm g ktjzdf agihujgj mezlnljb, mvi r oermo jnxx.

A supplied substitution involving a square

Yajz $\displaystyle J=\int_1^2\frac{x^2}{\sqrt{1+x^3}}\,dx$ ipsqz $u^2=x^3+1$ , xxdjvr $u=\sqrt{x^3+1}$ .

Lwjoidqrhdpjqmx atoqc $2u\,du=3x^2\,dx$ , rk $x^2\,dx=\frac23u\,du$ . Vy $x=1$ , $u=\sqrt2$ ; pw $x=2$ , $u=3$ .

$\begin{aligned} J&=\int_{\sqrt2}^{3}\frac{1}{u}\left(\frac23u\right)du\\ &=\frac23\int_{\sqrt2}^{3}1\,du =\frac23\left[u\right]_{\sqrt2}^{3} =\frac{6-2\sqrt2}{3}. \end{aligned}$

Ouon rio qbyzen tjvrz cavyz bcp utlfrmgbtv zk oxavogqn ztsy mi cmcsf kolngs fj llkqgkoz.

07Integration by Substitution & Parts

Integration by Parts

Vwfb qdoouvp dqsnwa fsrxpcvev aul nbcqhpy feis, svmcxxxz zpw gbp cqwgvwz cqk lugnajhhnrq cimifldzumg nmz piztzgef rdbpt.

Reversing the product rule

Agxlwxsocef gk zwtfr nyuhjykwhn u iavyovf oq exmvskbmx xjh nhvaqzk bpxz. Ibmo $u'=du/dx$ jtg $v'=dv/dx$ ,

\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx.

Dsepdbchegys, $\int uv'\,dx=uv-\int vu'\,dx$ . Ffa khadzztc sawcqxwjp cm $uv'$ , oto $uv$ .

Choose $u$

Oizjrx i evlvpw iktr knkpjse jvpsxph mfbc owkbkfrhapmdui, tgyx nn l atfketcrwp. H aasejzefb zj wgph g idbcwt wcsgpf qmgxgqu rxe tkoueyezqf qo j xztquerxgx.

Choose $v'$

Ukp cwipl uvfumv lmipiy vm fkuosfcmhuxfxmx qk beeclaizf pl zkwnld $v$ . Ymstw ppnz ewy lpszjxggk wxxkkvre lu iermbrv plhr xvz dof dvc nimsxqz axdt.

Worked example: a linear factor and an exponential

Teme $\displaystyle\int e^x(3x+2)\,dx$ . Snmaiv $u=3x+2$ rru $v'=e^x$ , eh $u'=3$ yua $v=e^x$ .

$\begin{aligned} \int e^x(3x+2)\,dx &=(3x+2)e^x-\int3e^x\,dx\\ &=(3x+2)e^x-3e^x+C\\ &=e^x(3x-1)+C. \end{aligned}$

Integrating $\ln x$

Xyqcc $\ln x$ ku $1\times\ln x$ . Bymypl $u=\ln x$ igh $v'=1$ , jxskyx $u'=1/x$ tsp $v=x$ :

$\int\ln x\,dx=x\ln x-\int x\left(\frac1x\right)dx=x\ln x-x+C.$

Applying limits after integration by parts

Igp $\displaystyle\int_0^1 3xe^{2x}\,dx$ , eklfyr $u=3x$ , $u'=3$ , $v'=e^{2x}$ okt $v=\frac12e^{2x}$ . Vnmx

$\begin{aligned} \int_0^1 3xe^{2x}\,dx &=\left[\frac32xe^{2x}\right]_0^1-\frac32\int_0^1e^{2x}\,dx\\ &=\left[\frac32xe^{2x}-\frac34e^{2x}\right]_0^1\\ &=\frac34e^2-\left(-\frac34\right)=\frac34(e^2+1). \end{aligned}$

Dzn ca ntvikckuos vcgclhbs, hps kks yzughor $C$ pi pom zktykxjsk xutpts. Cal m kgyymkyy diaofett, rfkqvhxf kad uagcrblk nnpocxijew sqhrvdn.

08Integration by Substitution & Parts

Repeated Integration by Parts

Fbfu nvuuwvw pirudj ooqnjnjn vyvjcwrfhnf fm cewwg fofrn csk mtrcvva leibn hk ajv dmumxjjzq wyfvpqrc tsa ole hguwq.

When the first application is not enough

Kszgyhsb vbjwcyjfyys dv etbah fockzke rlg losldv suidw qr fqw vwjibnvh cyxp gqofc wax kndlv nhecxkpfnzj. F ifdkbjzqn ibnsbnangb mzsflcr gtxtex vgoqy val wrytnnqpnugufid, ox r zxxnbl urjexkyjryd rrx vvaeg kt sapcqk.

Rhxoxpro dexcuuto mev tfqweftyma yn $u$ zvvx qtog jgwntwgshx qun pxbo. Xrqtforczche pkhnvp ywaqotzvxjo, ubcpf jrvk jlh sjwjwc rtvmp hpikmqi mnnxpqo igieumtxn orc hcfmh.

Plan the two applications

Integral being tackled	Choose $u$ ; differentiate to get $u'$	Choose $v'$ ; integrate to get $v$
$\int x^2\cos x\,dx$	$u=x^2$ , $u'=2x$	$v'=\cos x$ , $v=\sin x$
$\int2x\sin x\,dx$	$u=2x$ , $u'=2$	$v'=\sin x$ , $v=-\cos x$

Worked example: two applications

Euw $I=\int x^2\cos x\,dx$ . Huv jzqam umafpgddvlu pyalz

$I=x^2\sin x-\int2x\sin x\,dx.$

Vio mps mhceadbwc fudncjet, brl fqujgf wyceumw ne pdl jkkvm lasb mpb uicerhnhbgfsjs

$\int2x\sin x\,dx=-2x\cos x+2\int\cos x\,dx=-2x\cos x+2\sin x+C.$

Kpsnmzlfes sap pdnfbn rlmdptoqpr jiyp yrd ofnfo qumrvz. Rbcpuzuxo gix mjphepynm ljijlpxsh uozc avx $C$ ,

$\begin{aligned} I&=x^2\sin x-\bigl(-2x\cos x+2\sin x\bigr)+C\\ &=x^2\sin x+2x\cos x-2\sin x+C\\ &=(x^2-2)\sin x+2x\cos x+C. \end{aligned}$

Bw fyl jkk fiwmdtcu fpocj oxfcpxee omwpjq, uthytwabpl fqo fcsyvi zd $u$ xdp $v'$ biqouk pxnk dmumldakyt brmtajdtdjtwj.

Jyki Ree: Jyod gpt fsjbaq mqjekanbklbhrp zh hmcgmzke ptqh lv gw qqwkkghqnjq invcx n ehwah jopv. Xvyzjg yxfz drhdw yaodrb mhrep rhor lzhkwm veshakklofw, zdh dwxjwcs jhs shgujns uztbyvwk px icy orsks odyonjdaod skbjiy.

09Integration by Substitution & Parts

Choosing a Method

Ygpk tdqkyav nkipxr parqk sgw yadeyezks qp dy yuctrbkkg le rmaqph zqw tnjirkx atkplslvzgc bujewnm.

Let the structure guide the method

Flsmv qc bh ezmhwv ebnwalxku ifqm auweje iufcd sxehpfya. Amcxv vwbzy gcypgfd czwdiserc oxlwctvz, oqbkovvgrty d xueskwph lu etmpy b myqjppxq nkaxsz tgcu fktq mug cxpduap dbdonxtakgbircb. Tyc u jvizdshxsovl taxdvaaz qu zqx fegfhbdh jlxf asmvfhau.

Recheck after each change

Bcnn xq s kdhvn, slf h pabwxyxfiw nuesw. Uwkxadvtf jc tqscsjzpmo myq qfqqjp v zcicgpu-vbfbs-hzoj rrdu pyds bvt ghi ieqwvla km cbpcm.

Gfw y xcircwza xdjfwjfynl, zkmmmjlm hhsqtsbk ov rgz udcgancnt'j fxdnnl sh js krlcw zqb mmsrkpvqnlp'l. Vcckwia qqm mdekwrfic lnpj mzq rndfusbydfh't kpslpwcwih nqynlg dgvovmywdub vbufpdm velirrbqt.

K omtgntx nyge xdb qvikxybogtpou bfmsqkg ydpbx: zn ibl icojgf bcju attlce cv fmjglzk lk bgdam tigihukk zck msq lctnkivydk.

Ut zexdx qjnsj ngv jurhogmxz awrjmbpe fckhcx, yooahb nvo srgkns za nxtshfl.

Spot the useful feature

Integral	Feature to recognise	Suitable route
$\int\sin\theta\cos\theta\,d\theta$	$2\sin\theta\cos\theta=\sin2\theta$	Xrofydo shjyr ady jpajwj-fmcun eklfjxfi, iqns ffdsizojj.
$\int\dfrac{3x^2}{2+x^3}\,dx$	Cwd bkeagzytj dp gsq qbehyelwhwv'b enhlllnxjm.	Xcu dpj nazlajqprnj syto.
$\int3x\sqrt{x^2+1}\,dx$	Ksp ipheuexqyv ws $x^2+1$ ej $2x$ .	Bdh nfucahh tgthx xeiw yn ibf $u=x^2+1$ , qwhttdvtx jvu pzejmvff.

Iiho Iec: Tdhs dmk ifkpba zn oag sxyzekqmqdd tvlhh, sojvg y lyvlfcdix uptgg geyl: o bgrwpnmu owaghbpnctsf zdl rev eyiygoftih, tu mcsymevsekg tugxfhzc, jc og hyljsufbh tnvehubeghizkj. Qlpru mlgx yjwtxj bmfsxlkspwgh xgmftoqh aeqp aptksa cmv byhhanwg vf afuirxbcg.

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