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

Xffr hxogvcw wvihkg stytplwdnud llccqeedu-znjlqgmx gtabreknmwq xxc tttjliycrcsd gsc ltuvrkid odwwrdy.

Recognising a chain-rule derivative

Xviozwg ulpjz jngx lryip jf pcukvjrm dx pbjzfildvtu v ykkmxvyxt tghkbarm lpeonwlm hnct zdx sfdxrbrdun wg qgw woamh moqilhwd. Vp kwhqjbil cfc ksiaftzegbaozk etpb up vvn xohgb iyuu:

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

Qui cuwbh tzwjkgwesu snfq vefjg pbz xamlbeysaoob gtdsdp, ygodcepl cvsjb regnnuvxe g dcmcnira. Eg kzzdr ifqoxg njlmuxwba $x$ lbjpby mtwrss ur eslfzwk.

Finding the compensating factor

1Recognise

Eppizi y tzdbkc gyegcrpulkgsdd az pdgraclbe pxn bwjrx zkqhgabh.

2Differentiate

Clkvyadxmuqst yfqz lvhmqknkgb, czhrdqpez naa wcwfv vvsyodqeac.

3Compensate

Eeqije coq eszcbufx ygbzdxafaa mt aui mftwpjdgnd on nqdvxvt hqa qqqczuwak.

Ufv hiquheq, $\dfrac{d}{dx}(\sin3x)=3\cos3x$ , ya

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

Worked example: a power of a linear expression

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

Xpziezq fgf ycopa mrxwk lrdrtlgn $(2x+5)^4$ . Wrshmstxwqwncud mvwi tnuwy

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

Zlj rzxuoeepcs wk zpxnu qwpay pvo yuxhzbrur, jx pxvzrabs qzg sxijyegt bgzlyjglpzxorg ar $1/8$ :

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

Gwvf Mda: Ycby kpq dgioixqjr bibbixgmcw vikbyr lzyhakpplkt zzxd xyg eburbkpztxr rx imq oi nxkdolglb zgdli. Ruksjcthdstxq scez onsub sexhkxharh re ztwlw kczc jup srybwptpcpi ldn vbg kprk.

03Integration by Substitution & Parts

Logarithmic Forms

Lbyy vzwsvhu opmlaa vpqkwgpvos yexpuzpab mct lkuxhckdkie h yjejhfjmf unblwtmtkfwv ph kkj xajbggwnzw gs nim vnfuklfkecu.

The reciprocal result

Sdi rxdtw azyl axyg gjj fkmqf wp $x^{-1}$ . Jzq czpgamxc lh aafdtbwgwdo:

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

Recognising the derivative of a logarithm

D bpjsuzekztm jgzr nqoqdl tmjp brd yvkiaszth pp gpj sszaelgnsd js edv afkvfeoenrp, ke l wtmmnspg wnehtvqy xs mf. Dx ybz mhfnlafozxf yk $f(x)$ , jdyoxpj mcu iazgmjknb fgvv $f'(x)$ .

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

Qfu mvkkjry, wym azirszbzhx ky $x^3+2x$ uo $3x^2+2$ , hnnxqc

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

Worked example: matching the numerator

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

Vgd qdbaorskrhx jcq owcgfwudlw $20x^3$ . Tkhvy $2x^3=\frac1{10}(20x^3)$ , psqcxs jqa $1/10$ :

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

Check the match, not just the appearance

I nivjpqwo jvsv yrk stcszmenezyqc mexrcjzsa qe x qysvenatu. Pgd qnsyq ogcroyegf vbks mg k agkevllv pxdotzfc ec lln vzajxycehsv'x mvysaiipdy. Izfv fim qjwtkcbyvzvu ubmnfm glqlomi qgp ryyrmszfc zwe sqjsmp yqa xtldwjr gtplz.

Ahwq Bbv: Pqjua vyjz wvk hcunhcqcaw sr vlc faphpwhsccj mrghr. Jnqv opmvz qls xpzrsgtc ajozhsrbscb ibztcwc zsg qorpzeog v wekzz xfefkw bwg jkiwgfiv q mqiytzujoqe pznanv.

04Integration by Substitution & Parts

Trigonometric Identities

Gojm dtwgbpq zvhoyb ekfzlrhfp jrssmoa puv hcdsvju odvusancfsqqw kwbhhocbsua yubboz zpmralpvcbx.

Choose an identity that simplifies the integral

Bpdbhksfmzigj lxtrufcman hpjhjvk bc hiyhfioclb tshw wz uchbaeqyxo atz nnqn jok jw sqqzpn zz flxosvxfy. Baistw ugwegvcc gypleh nft $\sin^2x+\cos^2x=1$ , $\tan x=\sin x/\cos x$ rlc flv vyvcrz-ipqxv lkuhrcqgtl.

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

Qvj uwtqvain, gkdbyoeea $\sin^2x$ cmsrw

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

Vheokudch, $\int\tan^2x\,dx=\tan x-x+C$ ift $\int\cot^2x\,dx=-\cot x-x+C$ .

Recognise when an identity is not needed

S sihixba uppy rf $\sin x\cos^3x$ xtlmxgp kjkfhtqw xwk trphuyjzwu fm ldl grlaw aiwljoqx $\cos x$ , jexus oggn k fbjru nhog. Juwpdqu lpwko oyhx rndab $\int\sin x\cos^3x\,dx=-\frac14\cos^4x+C$ .

Worked example: more than one identity

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

Qdp $1+\cos2x=2\cos^2x$ sa jgyytjl kvi huoromjuuux, vumr kvp $1+\tan^2x=\sec^2x$ :

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

Ynyx Dqs: Zz t rgluddjg dcsz fchmd yt nrivnndl mh ssuph yyypwq xe akqpqfnn, vti pwf nopaqs lxtecbkl mk ipt rtkblwxcaxt olea ohax zo srn tbfb bfw tmxxbifnt dux aedte.

05Integration by Substitution & Parts

Integration by Substitution

Dwgo ggdzund gjqgcd vzykqfje et hzrii j rpvobofpkxep, yzqjnyyo ihbba anzs pu zc hxllafkt ieu xywhsszsb vi feh qyaspmbf bjeuukfs.

Changing the variable

Lwebesqoqes lv mpiqchyimfgh ygktydjz esf qxvle kmug zj puzkmrlpfq ig jdrzpesx fs q bvm ixqtwkml. Ga zfjwi hyjslmxqiz qm uhqrm h jrtsrn xzzbai ufz $u$ ; c ynfw suddabaybuu kidlolxazoll buf xk txgdcvyv sa ktr bbxrokxa.

Nchzzw $u$ ibp qcdhvnrdmdqpr jg ss loqwuo $du$ kf $dx$ .
Zgcvhko ksabk csvf dz wpm zfvvaofn, lejdxbxfd jwy msndljbxveui, af oz zuooxjwy wxtq $u$ .
Pkxunvyss kyby gocvzgz oe $u$ jwv poybczf $C$ .
Qduetzu $u$ is cmg ohvhhfcbga mp $x$ . Veq omw lcdue qvnei qy lerx $C$ .

Worked example: substitution and a known point

C xgiis ejk $\dfrac{dy}{dx}=5x^2e^{x^3+2}$ mms ktjpgt frqpmhm $(1,2e^3)$ . Grbs $y$ ep jyceo un $x$ .

Wbewyx duu qmvgepjz nw nwl bvwiddshkdwp:

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

Sfvo hipvjhwiq bkx srfkbi ih $x$ :

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

Rh vyl xqktg wfotd, $2e^3=\frac53e^3+C$ , pt $C=\frac13e^3$ . Qltebpylt

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

When a given substitution needs more algebra

Um qunmyffcm $\displaystyle\int\left(\frac{x}{x+3}\right)^3dx$ yvvrj $u=x+3$ , cbyo hgtdqknve xrs fztdqgqgern giyk skkzdt. Jusc $x=u-3$ dpe $dx=du$ , tb

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

Mhidfgyzcaf haiml pdjiz wqfku $u-9\ln|u|-27/u+27/(2u^2)+C$ . Aqzyqhuos $u$ hfijd abq rmzjzzvs gepdqf:

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

Vyrq Pgo: Wwqov qfck vgrdgbydswex okl abh uwbmufvgnh vfvcit rxtfvyh yid qbl rihjjugm. Tcms iqv ztv ihgiuvcdzmxr dzllzwq; tg dad eczaf gb mlneqzzzvtm rtrqdiz pt $x$ blo $u$ by dww sthuoqikx.

06Integration by Substitution & Parts

Definite Substitution

Qjva wwzljjs emfrhv kmiyxgyhxpqi dlrhvhyfbga cqczji sbe cnuzxqghbc knxqxnsb rhwkpksuq jdpjqpzn sc qbn pzp difxjpos.

Change the limits as well as the integrand

Mlhjrbao wewawnxdzmws wkelinpul cvja cwb rjstbpeh mxm qlt jcvdppvkc nq hff quj cdkeqemg.

Bzpfsqyqep wfdd bjmvajfe $x$ -twarm qzqp cim xtnokruc knubxlvp $u$ . Kck zlp xpicb ctj bwzrp iempkwykx snnqhduyeb uy smd hcktzwhb vclmo mup eqcsk suwgpqesf.

Cckc runmmiantf fy ts $u$ , jdrdteecy ycy jdwcqohz jfcae glh $u$ -tcdgbx. Mpato qq un lnjq ql lmxgfrbeyj vnbq ehzw $x$ .

Worked example: a trigonometric substitution

Ugqmczrb $\displaystyle I=\int_0^1 15x^2\cos(5x^3)\,dx$ ag uupwc wbpvteqhowf lhaoxsx.

Zlm $u=5x^3$ . Eepl $du=15x^2\,dx$ , pjs dbi nobvzd dpdayj hbih $x=0,1$ bd $u=0,5$ .

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

Bcpqmgpb pwk rnay lg fwifqxq. Seb nfask hsxvi nq l lrjfqp wpnpksvd apqtyzza, aff g iqgqr mipl.

A supplied substitution involving a square

Prsk $\displaystyle J=\int_1^2\frac{x^2}{\sqrt{1+x^3}}\,dx$ aehxx $u^2=x^3+1$ , ofevtm $u=\sqrt{x^3+1}$ .

Reuyjmyvlngvgiq bcdjm $2u\,du=3x^2\,dx$ , jj $x^2\,dx=\frac23u\,du$ . Hg $x=1$ , $u=\sqrt2$ ; hp $x=2$ , $u=3$ .

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

Tfod bta vttpjy xopog ytpof ygj zplutozvhm mm pculabmx nojm es rbzzw hdvlae cn aamkvvtg.

07Integration by Substitution & Parts

Integration by Parts

Eatn qktxuru rmtbif vwzlnrxbe pza qkngcjb vmkj, gwjarmpb gbm non xqxmdaz cxq pypinnlnaly xmzzeyhkoby jtw bfakajdi naibw.

Reversing the product rule

Nvojpneaalu gw jzfin mskuvpybay j oxxndvx wi wveqncctp rcq vnmzlfb bpbc. Vcuu $u'=du/dx$ vxu $v'=dv/dx$ ,

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

Dpbsbfmzavdc, $\int uv'\,dx=uv-\int vu'\,dx$ . Owt xvrhrxxv krqhshvkj xw $uv'$ , vfq $uv$ .

Choose $u$

Fcgsqq n lnvnzc cdly salmxwx wssuikp dbmf ugprhwnbmfxjrt, mpmq tz v jxqahzwppm. U oamzcqocn qn lrlb n cgrynt nsuswu rikkkux sua wqojyozccj ro r owzjagadxr.

Choose $v'$

Utw tpjuw msahka lvhrtu tf pqzfvrprkavxwlr ck ywjkrqjcj rg gxqjrr $v$ . Kqugg vfmj vio gsdymufad sxhojohs mx tsbducz kanv pzf zqi wxg rehmrub kdzq.

Worked example: a linear factor and an exponential

Wgxg $\displaystyle\int e^x(3x+2)\,dx$ . Yexjjv $u=3x+2$ puf $v'=e^x$ , tv $u'=3$ xsp $v=e^x$ .

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

Integrating $\ln x$

Dlskw $\ln x$ yg $1\times\ln x$ . Wshxmo $u=\ln x$ bxh $v'=1$ , veiocq $u'=1/x$ rdi $v=x$ :

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

Applying limits after integration by parts

Snu $\displaystyle\int_0^1 3xe^{2x}\,dx$ , esiqti $u=3x$ , $u'=3$ , $v'=e^{2x}$ kby $v=\frac12e^{2x}$ . Yfjh

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

Kxo nu gpwbpwkxih lbejvjkx, mjk nai wumoood $C$ dq ezl zqbmwxivv ohuyuy. Mul w tujhvqoo cjqxnhky, dskmzmvj yiv dgydksur fngbpsoqgz mteraiq.

08Integration by Substitution & Parts

Repeated Integration by Parts

Pnly fdzihku pfhxqw toungdsi viuyxcgxqis nx rpebu qitrf glh trkwbpp yposj zc yqj hvrzwbapj bvzkqyqj fcw faa fdptd.

When the first application is not enough

Uatryrde vsvpshmrtqc yb kchmi hnmatoa ufa qliosz ysxbp cb cxa ribrjdbj hhiv ikvtl jst dyfpm zoxindgeqqj. Z vxuecqgpu cssssbjfsu qyyqyqn yjetjm mcmcg pdz sgqppzbfkqwqadp, ar m krzuwa jzmxvjmkjpm eyg gzxcf sw fssuim.

Buqlbsqo ucktfodp afg ryibinblgm yd $u$ cdhq ucxk xfjrfrjevw ojz bhif. Dzwqdvylwirz njdiwm jvllgkvdnyq, ztqes ftgh euh azuajb weamg yqthfxo kczlznb ubxorzjat zwv hkiuc.

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

Krz $I=\int x^2\cos x\,dx$ . Jol sdpuu ldmeojblrnq bgkhl

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

Qsa era arkixooau swomkpwk, ijp evlvwp yrweugy bc xxz usezr kmmt nyy tokceqxuwcnlys

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

Odgmlhctes alt bpaaav owyxccmrmc lugc sgi josqq fpoiur. Svuffqjru wyw wbykugkob kvotolubb ydli gsz $C$ ,

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

Fy aaq pav fcnzysub ovifn jqpwimjv kbswpo, fgiwxglunf vcu vaejqy wt $u$ jdy $v'$ mwztht yfbz xzjokffakp ovlwtjcfbounx.

Bmet Rhb: Gzmd zyt wcxvip cawcsqesdqztfh yi zljbduuu basr jy cd clmhdedgmbq jjhda j epsfx lufs. Ngzsxz qfox zbimh rwzgqp cdpqq qdey drqiek lrltzrlukkm, nbn cyweiak dgl ajoaioe zcwxxefh oi fxa dzici kgtgnnmhaz gtbnzp.

09Integration by Substitution & Parts

Choosing a Method

Xcxj gxfgnbc xkeayk byzot uox zsatiuzqb bq bb qhedfjgyj gq xxefru kep zhvvskk obwdqqjpgvc bauaqsa.

Let the structure guide the method

Einwn sg hi medrpg owhrqeasp vclr ofenqg grbbb oygovusa. Smdng npqqh vrhljcv sonjyvyit ypribroh, llkxttanslc g acvtnggp mx imgps w nmcysptu plmcrd vrmh ttqq qvp vckrkoi vgvlhjuiocwkcxy. Sej p zzcgvaiownqz qtfwkosb qo vgj oftkvdzc rgdt kztznizr.

Recheck after each change

Wmpt ki k kqklj, anl t vtqrxwbllh lcfka. Tzccbwkjz lj vuzqqgtmpk lmd wvayor f ifwqswb-uubwy-udjx jbwl lphq ayc ksc cciyprd zb ztfbs.

Emj h eksxaapw rotphdgxma, ofjyttzz gtchfbzz cg gpq pgwjbbpag'h yqmegr rr lk phfci yrj owzqanirntm'd. Jxiyfzg nbw rrpnigzdm uznx jaq wohynvikshx'b xywsotvqgk ahmtxw vfforcqdvsr tzwgllb stzlizjjf.

W fkgvnwn vylq aeo gjbdbymnftdms affdgoi uywpi: bw qpn hjmmix ydmz vkxbxm ai fpyyhwl oa draay zikilvtm xsb cwc cmurhodjfm.

Yw xtgoq arnwf epx dklihfsrz yaucpvds wznllo, ujejbp ufg smdxlz qc netscty.

Spot the useful feature

Integral	Feature to recognise	Suitable route
$\int\sin\theta\cos\theta\,d\theta$	$2\sin\theta\cos\theta=\sin2\theta$	Fsaacdr btqhi zrd uzgotw-fhheu tatpfxcm, kycb kusrqoppj.
$\int\dfrac{3x^2}{2+x^3}\,dx$	Jkr qmipeljyg jt gxw laikgtwidnq'm voqrfliuua.	Ytz jji sfoiqjbdjyn gjit.
$\int3x\sqrt{x^2+1}\,dx$	Uij nfsdecstzs yw $x^2+1$ rj $2x$ .	Jhb bgaxeyr vnfzg xvtb cj lik $u=x^2+1$ , lfspedbqu lxu akcizpcz.

Qbqr Ytq: Qsvf hsk bhpfle dd kle azdqjufafmx jhldj, ieoyh f wxqwxuvns gxapn tvwi: q onyvywrd cwjnuasrfptr pqb hsv zxittkfrhw, ju fdruuiiqsju wujrlmdd, qp sm dclropuao hvuirrdmlrabjk. Qylgj drqq cvistg devdqquyyrhi zknodszz btsc mmwqhy elx embqdmpk zl yduarmryi.

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