Integral CalculusIntegration is the reverse of differnatiationFor example -d (sinx) = wox/ CDSX dx = sinx + CdxarbitraryconstantIn indefinite integration, no limits areapplied and thus an arbitrary constantis added to the result In definiteintegration, limits are used and theintegral thus represents the areas boundby y = f'(x) the x axis and the linesx = a , , x = b, , if f'(x) is continuousin [a, b] -aa(f'(x) dx = f(x) = f(5) - f(a)bbThe area covered is shaded in thefollowing figure -+ve areax = ax=b-ve area In an integral(f(x) + g(x)) dx =f(x) dx +g(x)dxIk.f(x) dx =kf(x) dxA fewofare -(...) dxresult(...) dxkkxtanxLn/secxlxnxnth/ntlcotxun/sinx)xln/x)secxLn 1secx+tamiexex=Lultan (x+)axax/InaCSCXln/cosecx-sink--cosx= Lultan(2/2)cosxsinx1/41-xsin'xsecxtanxsecx=- coscosecxcotx-cosecx/T+X2tan'xsec2xtanx-cot'xcosec2- cotxTxirx2-1sec' xCSC+Mostintegrandsofsubstitutionexample(ax+b)"dx=?Lett=ax+btndt/a1/aSt" atdt = adx=1.(ax+b)"++ana(n+1) DatePageSome reverse integrals product can rule be formed usingForexample,f(x) + xf'(x) = x.f(x)diff f(2)(product rule)diff xTo integrate the product of two functions,we use integration by parts -Iuv II dx = ufvdx - dx + cIntegrate this fune ; sometimes V=1differnal this tuneIf the integral is definite, apply the limitslike so - -bbbwdx = [ujudox] -(u'/vdx).daaaApplying integration by parts repeatedly lookslike this -J wvdx = uv, - u'v, + 42V2 - u3Vv +(until un = o)some integrals can be formed usingpartial fractions For exampledx = ?(x-1)2(x+3) Let x2 = A + B + C(x-1)2(x+3)x-1(x-1)2x+3because its a repeatedAx + Ban unfactorisable quadratic,root if it was(quadratic)we would writeThus x2 + I = A(2-1)(x+3) + 8(2+3) + ((x-1)2,From this equation, we can comparecoefficients to find A,B &C.Putting x=1, 2 = 4B - =B - 1/2Putting x = -3, 10 = 16C - C = 5/8Equating the coeft of x2 on LHS & RHSI = A + CA=I-C=3/8ThusI=f3/8 +1/2+5/8x-(x-1)2x+3=3 In x -11 + 51ml x +3 - 1/2 + C88x - IThere are several common forms ofintegrands that can be resolved asshown below.For a reciprocal quadraticfdx=I-tan (x)take a commonx2+a2in denomafdx=1.InX2-a22a/x-a-dittox +a PageFor a reciprocal underroot quadraticdx-sin (x)take acommonVa2 - x2dx=m/x+Vx2+a2x=atandx2+a'dx=m/x+ Vx2-a2/x = asecoVx2 - a2Fora linear quadraticlinearorVquadratic,rewrite the linear expression as afactor of the quadratic's differntiation,thensubstitutethequadraticastIfthereisasimilarformwiththelinearinsidearoot,substitutethelinear as t2 insteadForareciprocallinearquadraticfI . dx= ?(linear)" (quadratic) 1/2Substitute linear = 1/t or use -dx= sectxrx2 -1The same technique can be used fordx(ax2 + b)(px-+q)" For a linear quad. Vquad + substitute thequadratic under the square root with t2Foranunderrootquadratic,/ Vx2 dx = 2/2Vx**a2 I a-1 mlx +S Va2 x2 dx = 24/2 + 02/2 sin (2/a)Foraxesinbxorcosbxdx,e ax Sinbx dx = e a2+62 (asinbx - bcosbx)oaxeax cosbx dx = eax /a-+62 (acosbx + bsinb2)ForSf (atchtd) 1/(cx+d) dx, writeax+bcx+das t. . Thus dt = ad-bc (cx+d)2andtheintegrandcanbewrittenas1/kSf(t)do,where K = I c a al = ad - -bcForreciprocaltrigonometricexpressionsdxmultiply - divide by Va'+62asinx + bcosxand letfactsso6adxditto or multiply-divideCasinx + bcosx)2by sec2dxconvert sin and cos top+ qsinx + rosxhalfanglesintermsoftanfp+ qsinx + rcosxwritenumeratorasa + bslux + ccosxAldenom) + B.d (denom)+ C For functions of sinxcosx or sinzx, writet = sinx + wsx So dt = sinx It wsxthenwritesinxcosxasintermsoft.For functions of seex + tanx, writet = seck + tanx so dt = seex(tanx + secx),secx . dx = dt/t if the integrand doesn'thave an extra secx replace it with1/2 (t + 1/t) -Secx + tanx = tsecx =Secx - tanx = 1/61/2 (t + 1/t)Forfndx =Suztanx.tanxtan'x dxIn =tann n-2 x sec2 dx - / tan"(ne W, 43,2)In=tan - In-2 2n-1similarly for secrx dx ,In = tanx . sec"-" + n-2 In-2n. In-1and for (sinn x dx,In = n-1 In-2 - sin x cosxnhand for f (cos"x dx,In = n-1 In-2 - cosn-x sinxnnFor S sin"x. cos"x dx and S tan" x secm dx,etc - split the odd power so that you have Stat or similarSome integrands are twins andcan only be sowed togetherFor example,fdx&fx2 dx1+x"1+x4(I,)(I2)I, = 2 1/2dx = 2 (1+x")= 2 + 1-xx) 1+x4= 2 / 1-1/xx(2-1/25*+2 (x+)/x)*- 2=2 12I ( tant(2-k) - I In / +1/2-52 1)F22-2x+1x+vzI2 = can be found by performingsimilar partial fraction nagicAnother example is -S tanx & Strote(I)(I)I, + I2 = I = tanx - + Vcotz = f sinz + cosxput sinx cosx VsinxwsxtI, - I2 can be sowed like that too.Thus. I, = (T+Is)+(I-I) & I, = (I,+I,)-(I,-I)22 DurePagesome integrals can be solved via substitutionafter manipulation For example,dx=fdx=x2(x"+1)314+1/24)-314dxx5(1+1/24)314x5Let t = I + 1/x" - dt = -4/x5. dx Thus,dt=-t 1/4=-t3/4--441/4(11/11/44+Cx4Another such example is -DOoluxdx=f+in'tde if x = 1/tax2 + bx +a(9/2++ b/t +a)t2dt = -1/tz0008I=-Lnt dt = - Ioa+ bt +at2Therefore, I =0 DatePageMost definite integrals can be solvedusing the following propertiesThe variable within the integral canbe replaced with another one withoutaffecting the integral For example6bf(x) dx =f(t) dt =aaPutting a negative sign in front ofan integral will change switch itslimits -baaf(t) dt = - b f(t) atA definite integral can be split like so -rn/2rtank dx tanxdx + tanxdxn/2More generally a C f(t)de = a b +In the graph,f(x)fla+b-x>x=ax*bThe area under the purple and blackcurves is equal Thus, - Page6= [H(a+b - x)dxThis property is also known as king'sproperty, and the next one is called queen'sIn case f(x) = f(a+b-x) the graphis symmetrical about x = (a+b)2 Thus,more generally if f(x) = f(c-x),cof f(x) dx = 2.1 f(x) dxsince the graph is symmetrical about c/2.If the limits are symmetrical about x - o,af(x) dx = "[((((x) + f(x)). dx-aif f(x) is even, f(x) = f(-x) and if odd ,f(x) = - f(x) so I D in the second case.If f(x) is periodic, f(x) = f(x+T) Thus,I{n}nT J f(x)dx = [n]/f(x)dx + f(x)dxTat "(f(x) dx = (u 4-m)./t(z)dz if (n-m) E 2a+f(x) dx = = zeo if f(x) is odd.Note that when the limits are reciprocalsof each other or one limit is 00,an appropriate substitution is x = 1/6 or tano PageIf f and g are inverse functionsof each other,g'(x) =I& f(g(x)) = xf(g(x)) or g(f(x)) = xAlso,a by f(x) dx + f(b), g(x)dx = bf(b) - af(a)bHa)Somedefiniteintegralshavememorablevalues For example,nty Ln(sinx) dx = n/2 Ln(cosx) dx = -T1/2 ln2n/2,0(I,)(I2)n/2r/4I, + I2 = In (sinxlosx) = 2/(1m(sin22) 2 - Ln2)/I = 2" In (sint) - dt - 2.1 n/20 Ln2rd2xI =I-n 1n2I = - T/2 Ln224n/4Also,tan n dx=In=Io- In-2n-1For even n, In = I -+I-nn-1n-3h-54Foroddn,In=1-I+I+ Lnv2h-1h-34-5rIf In =sin(nx)dxIn=0sinxIn-2 DatePageSo for odd n, In = I, = n and for evenn In = Io = 0For such series, always check In -In-2or In - In-1 to obtain the reductionformula,If In = of (1- - xt)4 dx then In = kuIf asked for In, tellscope - In-, (kn)+1to get In Im- 1. K1 I, = km k(m-1)Im Im-2 I, Io km+1 k(m-1)+1n/2For sin"x cosm dx , , the value isk[(n-1)(h-3) (10r2) [(m-1)(m-3) (10r2)](m + n) (m+n-2) (1or2)where k = 7/2 if m and n are even else k = 1.This is also known as wallis formula.For definite integrals the newton Leibnitzrule states that -v(x)dx d (j f(t) do) - v'(x) f(v(21) - u'(s)f(u(2))u(x)An especially useful application of thisrule is while apply L'Hopital to findthe limit of an integral For example,lim /sinrEdt = lim 2xsin/21 = 2 Lim sin/x1x-o x3 x-> 3x7 3 x->0 x= DNE Similartodifferentiation,integrationthefirstprinciplecanalsobedonebysuchLet(xx)beanMIfunctionthatf(x)>DforxE[a,b]To findbf(x) dxwedivide[a,b]intonequalpartsofwidthh = b-any = f(x)5m = +Cax-p(inside the curve)Theareaofarandomrectangleishf(a+mh).Theareaofthebiggerrectangleishf(a+(m+1)h),Forboth-theseexpressionsmstartsfromzero.Thus,sn=Ehfla+mh)while Sn = Enfla+ rh)bandsV(x) Show description