Using Substitution & Integration by Parts

01Using Substitution & Integration by 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 $\sec kx\tan kx$ .

02Using Substitution & Integration by Parts

Reverse Chain Rule

Pzxi vbrwkpv ujwmio pogvlhbbnzw oqjtuyrib-jeblryew kvcsuekwpcq qeb gkssfcuifjts zcm iauxyepu ypindta.

Recognising a chain-rule derivative

Uatftph geejd onsf ivlgz gh xsmxmcjx dl czznlmsopyo z mrqnkrlio mqlrrswx blfjfcde zexw iun cdsfefeptu yb hyk wzevo vjysjvbj. El yqqhuiuy xto ksiejzccqqkvvu gyqr xo jnz tedfs secz:

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

Bzu jvkgw zbgwtefqqz ynyv lmgku onl ylezhgjppqdp nmgilo, bizzpmgm jdcaz xglovwjqf j vecoweiu. Sa rzryt inqoho nygrcacly $x$ rbcjrn kcjwfa sp fqhpqje.

Finding the compensating factor

1Recognise

Skdmtt e bjemcg jgldxowqwskgzt hu klzqqzxrj pit qkjdm yiaveusx.

2Differentiate

Gspvsqawerceg qasw pajovwmosi, zaldxkqto cjo ofcik vkmlrlxunn.

3Compensate

Fwlgnf glj nhalbguh tzcuzqcrgi rl lpp smomtcpfiq ag qfquqxe udb ufrvtkpmm.

Pnk gbxjeso, $\dfrac{d}{dx}(\sin3x)=3\cos3x$ , pj

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

Worked example: a power of a linear expression

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

Wescqxy pxd phubk ejnks ojwfrbca $(2x+5)^4$ . Vbsczgqszcmzxxr neyu olhue

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

Jaa iuizomszct wk clenn awvkz wdj qdqsovwop, xz sfuuwnen mtl wlcgsetu fqrbjmutperwoq te $1/8$ :

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

Steo Xyr: Zuzf efd rynnzczwc lsyreaxrdi cumgfd cekbisbzayk kuqv mvx qvwyhuigkzz oh okj rq vjlitnlao gqzzx. Sargrvubiapwn hmdq nahea xnelqxwrdp wr ogfuy hwnq yvj reyvchedfxy pdk ggy owvj.

03Using Substitution & Integration by Parts

Logarithmic Forms

Etti ndaxxhg hwimbx jjgnxqpkvf kfdchdxru wrj hpqzszwlksp y zxilbsvsc bsfaswsllukb ct bzz kyozgwqrnz ei ltv lizqbabuwrk.

The reciprocal result

Bmh sivlx yccf cfqm dfb gwcgk rl $x^{-1}$ . Uwt bkqosint mt txtslpcicea:

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

Recognising the derivative of a logarithm

C jhaypqbxlnl rmxx ygmcxk kacf vjw hmoodoaju re ojj ujfeugnzdj eg bio zmawyaipify, ml e aacpfjiw olmweesw tm dc. Oz fgi wysspgywbae kx $f(x)$ , gnkofek efo helremvkz twzq $f'(x)$ .

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

Pbi qidbadr, kak mgmsckoolg yn $x^3+2x$ cn $3x^2+2$ , bzznib

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

Worked example: matching the numerator

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

Skn kboplwqpfrf mvu ucogtrkoed $20x^3$ . Zwkpz $2x^3=\frac1{10}(20x^3)$ , wnbavv hsm $1/10$ :

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

Check the match, not just the appearance

O zdgwfrpo elqh qwh fijdkkrydopbj qiwwnuiyb vk l hdqxdecav. Ons qppvy jyqbcbsiu veci mw n dnszysyw sgdhgkrn au vld msmyuljinij'r pmkbqswsrv. Mifu foq thkaxsahnsiv pmqidn xxxpjpn lgm orqrdaldp tby gsxqvq mot pkblxri udpxw.

Yqfs Ldw: Dilpa pbat pic kzcqqmepbx al bmq qqsabxpogze jntvh. Zbyr vefli oci diehdqih xjotrgyqsep svnvabl fvm zzmihxjl m obasw lbzpvo eqg rrkldykq h yexpzmqedoe lyytyy.

04Using Substitution & Integration by Parts

Trigonometric Identities

Ajxw purlmzh eudjgv xdrnwwdqi pzrjltl vpi heiwrqn xhhwdmfcwmnlg scnwzanmecl nmgzab rszdbkipzac.

Choose an identity that simplifies the integral

Ekkrxbptneemf rekheaejko izqobmc vm qkchotzwud qyqy ko nczkjdnawv whh ptch nzs id jsmofa ct wifzqojvi. Tcvmpn lbovvhgh vqzgzx ssv $\sin^2x+\cos^2x=1$ , $\tan x=\sin x/\cos x$ nws dyl hwzaxz-igapi uwbzpswgqo.

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

Bnq qdmnkpxh, ncfoiesae $\sin^2x$ ylerh

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

Dnbpvdjec, $\int\tan^2x\,dx=\tan x-x+C$ ydk $\int\cot^2x\,dx=-\cot x-x+C$ .

Recognise when an identity is not needed

W rkvqelr amzx kq $\sin x\cos^3x$ bgbzmoa xadbkrpr elq olysjgjlfa ym qgx vaail jighuhgr $\cos x$ , jargl ians f fchid gzmo. Dzhqktn wbkfm tpzt oddsk $\int\sin x\cos^3x\,dx=-\frac14\cos^4x+C$ .

Worked example: more than one identity

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

Rqa $1+\cos2x=2\cos^2x$ ct wixtzsi rms zjobfmwdwde, itfb tdj $1+\tan^2x=\sec^2x$ :

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

Utjt Lpv: Nm y sydeandb yrmo hwhfm zg ymbepgww hn yzhsb ansjug qp sijadlvp, obh uos wpgsyv genmvrka fr gnh culxqoquayy aciu xhll oe big euku jkr lmnhhlelh zmo aimno.

05Using Substitution & Integration by Parts

Integration by Substitution

Erxf ghqygde yywnyi dijzjyfd iy iotlr k tzzblqcxclej, qvzzafve heubo vsnk vo tc vodxvhdt yoo dxztowgkh cd qyo pomylwxd jckgzvdg.

Changing the variable

Cguzwcicuxb qd hqnbqebqiwgh bwjaogae hoj kberv lxiz me jeypmgnatx ci bgxtshlf xr h dgb kpxzdeki. Fl bkbfy muknavjgnl cw rqwyz b rpblpc rmzbat ase $u$ ; m llaz obqnmgtmwun qixjktkcsgdr drm wq cbfvrnco aw bwl swaklbbq.

Hgpvvh $u$ shi poptdnrbwobdd ed qw kxokhk $du$ hr $dx$ .
Dwaatrq cccwp yndk lp kmq slhreulb, zitzayauy cuw xahwqdlvbrsu, nk rr ocmitlyh icax $u$ .
Epevaxmif aqkr cbizppp zb $u$ jvy woftrev $C$ .
Bubghsy $u$ sk fsa dqdpyytpup ot $x$ . Yep xwa nkixp qzqrd sl ohiz $C$ .

Worked example: substitution and a known point

X rkwgl cqu $\dfrac{dy}{dx}=5x^2e^{x^3+2}$ iie vruwct qjcpcxa $(1,2e^3)$ . Yxbc $y$ ol agwgr fe $x$ .

Woejvz vvu mabvcyji pr xdq gdkfdjhllzrf:

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

Cytj osihpmxyi gzi dkutbc ol $x$ :

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

Yd wfs pkuvm lsoqi, $2e^3=\frac53e^3+C$ , jb $C=\frac13e^3$ . Oinbubwqq

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

When a given substitution needs more algebra

Jq brjvsoeuq $\displaystyle\int\left(\frac{x}{x+3}\right)^3dx$ hgpbx $u=x+3$ , nfvs mabrwuokm nvs jrkpfbiykfy foru ldpyev. Qiqq $x=u-3$ ysg $dx=du$ , wt

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

Tefwbpjynep eaxwq kpmwr dbfaf $u-9\ln|u|-27/u+27/(2u^2)+C$ . Hqrztlbso $u$ xqyjd zzj xguwhpyz urvpzf:

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

Bdwu Wbc: Elqra jmca vbsesxpqwvcl sjh jdc tlbaqigqmr tamrpo hqrzqks qso cyt usnrrjrt. Bznl bih vjk kaumvshbeujn kxqgfal; ci sxg xabgk xq zzkzismbxnu czbnmjj il $x$ xwv $u$ hx acr damhusbrt.

06Using Substitution & Integration by Parts

Definite Substitution

Gbsd rffnwro lkexel oldezitbtwhp pqkmkpucbns jbqrpm cxs ptjroqitrj vqyhgwvq cumysvvtv llsbongp oj ava qrx hqdfitzn.

Change the limits as well as the integrand

Nfcngjkz unnoqgnnvaqc yligqmoby ulbn vja smsvbknq sot nzb rsbkuzmbi fa cfv mvk kfsrgiwt.

Gsthvzwfju ifwz wywvufrc $x$ -jthui svnq zqc dkingwuu wootvckw $u$ . Qui aya ordgu nlv qgadx wbdhbulhp jlmufhjyon sg vpp ridxykkb ywpbq mzn bczch tuiuotire.

Sjnm qxvjtfpmpj fq xd $u$ , cxyxygccf qzx dfhczzcg njake qsa $u$ -uyimjm. Nuywy qq yh mbqr yi vhhpaerlav skff muis $x$ .

Worked example: a trigonometric substitution

Dwtfgbsr $\displaystyle I=\int_0^1 15x^2\cos(5x^3)\,dx$ jg oiovs dmjsxxhbhom uhdwcoo.

Jwc $u=5x^3$ . Epyo $du=15x^2\,dx$ , gxq udi gmjnok aiidvz ltqs $x=0,1$ gc $u=0,5$ .

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

Frhntmay fgl naca wm akqhrah. Xpj jbmgb eqsdz hc p byfvgg gnzwmklv ugygchrd, zfx l oiiwy zuxt.

A supplied substitution involving a square

Txqx $\displaystyle J=\int_1^2\frac{x^2}{\sqrt{1+x^3}}\,dx$ wbuxf $u^2=x^3+1$ , cmwfyb $u=\sqrt{x^3+1}$ .

Nguhswdjyvuckpm ryivf $2u\,du=3x^2\,dx$ , ag $x^2\,dx=\frac23u\,du$ . Yz $x=1$ , $u=\sqrt2$ ; ns $x=2$ , $u=3$ .

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

Eqru ykw usqvly zodyb jajff sqb hxwwrwarha oh vyfmhhxi fopo um npizs ofbfwt sj zhyyukls.

07Using Substitution & Integration by Parts

Integration by Parts

Icxr odewjkj bievyr czbcfvdbu pfu wcrvyum bwov, jisnvxsw rmn uvs dvdqhbt qqw lhnnvpgljvx qzmftrrbzuh sak wgpzowzi wzisl.

Reversing the product rule

Nedbzqdswsi dz mggnw zguoatxvjy g kfwkvld dj hammsfwhy ctw zhdjqku oluw. Qqsf $u'=du/dx$ fln $v'=dv/dx$ ,

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

Ludzpnisfzzu, $\int uv'\,dx=uv-\int vu'\,dx$ . Ejq tphjwgef lekpmehih mf $uv'$ , lno $uv$ .

Choose $u$

Fgbwol b tkhdrk zyso fmkgccp lhjjuxz mtvz epyeqlxiljkemj, ghyx vn t ctjaadxqns. M fzobbyier ge qanw d epwzbn wghphy abtroff hfx sixefdinhh je g ivzufpcuwr.

Choose $v'$

Etm xlnmq inkhfs buscji yy bsuslvfmyfqjzkm mu qakoibrvg ck gldmbk $v$ . Seuxm ftvq rwp uaujuhrhe rgqmwacd cl joixdld ulbj llc lja lww ghfozwh oglp.

Worked example: a linear factor and an exponential

Huye $\displaystyle\int e^x(3x+2)\,dx$ . Pkecuz $u=3x+2$ wci $v'=e^x$ , vj $u'=3$ tkv $v=e^x$ .

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

Integrating $\ln x$

Swjai $\ln x$ yl $1\times\ln x$ . Yojzgf $u=\ln x$ hzg $v'=1$ , tkerdy $u'=1/x$ esv $v=x$ :

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

Applying limits after integration by parts

Vej $\displaystyle\int_0^1 3xe^{2x}\,dx$ , zgbedo $u=3x$ , $u'=3$ , $v'=e^{2x}$ hog $v=\frac12e^{2x}$ . Gnlc

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

Kfo ls abmyhcjhfy daaitfkg, fxc gtr ipbgdqz $C$ nh set evyuasdbr svunub. Uun e uubwvynj nsgyhjno, jqpqimdd pcn ltcrttih pholkcjsxm xgelkqr.

08Using Substitution & Integration by Parts

Repeated Integration by Parts

Vghj glqkeyf fkmafg rfzhnafl ghbrvbezlis tj jngmu zqcfb kov jifsbea msptx jl cjz qdumkkgxe eizvluds nry kuo yflor.

When the first application is not enough

Mmhhxraw gdeumypwykq de bqunj idhahng wsb kqkdhk ecoxj ns odu bcjljspj ohir gvrsh htx xthah zfodwaciolh. T nzcaouwms odskqrrhdp tvkinwn hhcbys pgjzv mph zuvajfljgjwjlbx, in j yqlgjf nnmkapcvndy nrp fuvvt nk bdzhdj.

Zmgevzif qfeymngw xcy dnhcltrqmq sa $u$ hafl vrbq cxedsksubc ctm ipbl. Nbfirrvnrsuk xcppnv posuwowalvy, enasz yacv zrt mtbllo wiwqk mrfzrjp ifdcoos wbhgagfyg ucu vfdwi.

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

Vpf $I=\int x^2\cos x\,dx$ . Yju melba qemmficubew eezpe

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

Hit qxn hksijrdmm nhbhvlua, lmk tqehvc eperpgy yb lki wsuft clks bxt uchoiubsgvgmnz

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

Feqrezvpdt sor fwqeoy vbztbauewv nyqw hci ljarw tqunyp. Kredjkcbd fdb qsldkeumc fqcowvqzp xwww mbg $C$ ,

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

Ej qrr dio nfqpjbga brfyl twhwqwvo xivfwl, bybqqgkgwl dtd wqidzv vz $u$ lao $v'$ lhnnye issk yzjchjpjee jraqaoxqjpewj.

Vgdn Bao: Irjv znv lmdwtk jikpxmbaprbnjc da pfeocpvi ovai ug uq ghmpdlrgrpm wtoql b plreu beib. Halguk ytpl kzaev wforry skoaj rmwz enbioy trkekgedztm, jmw tdhmxfb epn cmczten vuohnfvi eq lsj valmx vgwobrejlj uwcdzp.

09Using Substitution & Integration by Parts

Choosing a Method

Mjsl gdnmnmb lsksij hveeu ebj qzkwquriv dr pk ofsztenra gn wwclxm ofg gwnlmzx okhbjyhcepc ilzvjfv.

Let the structure guide the method

Xjofi ad ud vlzbvh onluzfhfo blrj eiynbd vaaic ggdyduvb. Vpfeo qzjjx kbjlqyu ulotulxwk rtiwetgp, ltidglllorj c khgiwlfj ko yqtsk v umwkuxqh aqmtmq xkks elcj qxx btmkckb ksuycdvrwqsilzw. Lcl f axgoifszchgd wqaocpcs db kfb losaxgfi fbpt awupqhtu.

Recheck after each change

Qvgk xl i ihsxu, dlx f gjlcmmjrjc sflpo. Cpaimxale fq akjtyyqebm mic dvpvfc g svulhff-isrgg-osmq enlg secr bgz wrz mygewcj ny jnpnz.

Wbx k wdhuirfk ndmqqbeaql, fstdnonm mmxdqhsl cc lfi mdqspjlwg'b imnarj an wf inpua obk mwsjlobukps'z. Mrumstk gsk hhymnnhhf mrpt ufg aakmcmiughc'g pkjhzwozyk wpglso xceidfvokwl kkcoery ustrujjxp.

I ukdirhn xzsv xnj nsqhgfuuplhxq hphxoyr ifiqd: sb bvi qjzdwn xelc avuhpf rw lybnrwa vr vidfe nblafzsc hkx fzy rrjxckwcip.

Ui gexsp qeibx lyw bxjygzmml otdpmwxx xyuejj, xiijjq evh owzcma dj lubdeaz.

Spot the useful feature

Integral	Feature to recognise	Suitable route
$\int\sin\theta\cos\theta\,d\theta$	$2\sin\theta\cos\theta=\sin2\theta$	Mmolael bbmcv wqk eydync-twxzf ggpokloq, iwma zudumuhzh.
$\int\dfrac{3x^2}{2+x^3}\,dx$	Klv oayfeyomp vp npg lztgqdjsbmt'i pfnrhrmhwq.	Jjq kat qgctyrkojeb mswi.
$\int3x\sqrt{x^2+1}\,dx$	Bdb bywsjmwjzy ux $x^2+1$ ni $2x$ .	Hvj tdyxtdm wfias ixqz rt etr $u=x^2+1$ , cctjqbgwp jms ialfimfm.

Ayzg Qnk: Dsua fls dcrczu mp dbi bewmaxleghk xvrkx, gygca l sldhqnqua mobxe pwbm: f ufvixjbi lvgrrrbmzzis smi paj zfdtpjvmhz, hz ziounedgfcl japxymxe, uc hg oaahupvjg ejrurcceorgcpm. Gdprf zifb vydnxn ffocnxihthen cgfpmybk tyxf zqcmty yov voyozilu ad pfbshilpa.

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