AP Calculus—Integration Practice

1 AP Calculus Integration by Idea:Ifu=f(x), thendu=f (x) have xdxx4+ 1u=x2=dx= 2xdx12 duu2+ 1=12tan 1u+C=12tan 1x2+CPractice Problems:1. x3 4 +x4dx2. dxxlnx3. (x+ 5)dx x+ 44. In each integral below, find the integernthat allows for an integration bysub-stitution. Then perform the integration.(a) xn 1 x4dx(b) xn 1 x4dx(there are two very natural choices forn).(c) xn1 +x10dx(there are two very natural choices forn).(d) x61 +xndx(e) xne x2dx(f) xne2x5dx(g) x5 1 xndx(h) x6 1 xndx(i) dxxnlnx(j) dxxn(lnx)7(k) xnsin(x6)dx(l) sinnxcosx 3 + sin4xdx(m) sin3xcosx 3 + by Parts:Basic Idea: udv=uv v du(Try to substituteuso thatdudxis simpler thanuand so thatvis no more complicatedthandv.) have xsinxdxu=x, dv= sinxdx=du=dx, v= cosxdx xcosx+ cosxdx= xcosx+ sinxNotice that in the above, settingu=xyieldsdudx= 1( ,du=dx), which issimpleranddv= sinxdxwhich givesv= cosx, which is no more Problems:1.

2 Xe x/10dx2. x2e x2lnxdx4. xnlnxdx(nis an integer)5. x2sinxdx6. x3e x2dx7. x3 x2+ 1dx8. Assume that f(x)dx=g(x), that g(x)dx=h(x)and compute(a) x3f(x2)dx(b) x2n 1f(xn)dx9. sin 1xdx10. (sin 1x)2dx11. tan 1xdx12. sec3 d (Hint: writesec3 = sec (1 + tan2 )and integratesec tan2 byparts.) Idea:a2 x2 For expressions likea2 x2substitutex=asin . Thenx2 x2=a2cos2 anddx=acos d .a2+x2 For expressions likea2+x2substitutex=atan . Thenx2+x2=a2sec2 anddx=asec2 d .x2 a2 For expressions likex2 a2substitutex=asec . Thenx2 a2= tan2 ,anddx= sec tan d .Example have 4 x2dxx= 2 sin =dx= 2 cos d 4 cos2 d =2 (1 + cos 2 )d =2 + sin 2 +C=2 sin 1(x2)+ 2 sin cos +C=2 sin 1(x2)+12x 4 x2+CSECONDEXAMPLE. In many integrations involving a trig substitution, there is theneed to integratesec.

3 This is easy but requires a trick: sec d = sec (sec + tan )d sec + tan u= sec + tan =du= sec (sec + tan )d duu=ln|u|+C=ln|sec + tan |+CIn an entirely similar fashion, one shows that csc d = ln|csc + cot |+ s one that uses the above ideas. a2 x2dxxx=asin =dx=acos d a cos2 d sin =a (1 sin2 )d sin =a (csc sin )d = aln|csc + cot |+acos +C= a2 x2 aln a+ a2 x2x +CPractice Problems:1. 9 x2x2dx2. dxx 1 x23. dxx a2+x24. 4 +x2dx(Hint: see problem 12 page 3.)5. dxa2 x2(It might be easier to do this by partial fractions.)6. x2 a2xdx7. dx(a2+x2)28. sin 1xdx(Letx= sin )9. (sin 1x)2dx10. tan by Partial Idea:This is used to integrate rational functions. Namely, ifR(x) =p(x)q(x)isa rational function, withp(x)andq(x)polynomials, then we can factorq(x)into aproduct of linear and irreducible quadratic factors, possibly with multiplicities.

4 Foreach power(x )nof a linear factor, the expansion ofR(x)will contain terms of theforma1x +a2(x )2+ +an(x )n,wherea1, a2, ..,anare all real constants. For each power(x2+ x+ )mof anirreducible quadratic factor, then the expansion ofR(x)will contain terms of theforma1x+b1x2+ x+ +a2x+b2(x2+ x+ )2+ +amx+bm(x2+ x+ )m,wherea1, a2, ..,amandb1, b2, ..,bmare real determination of the constants above is a purelyalgebraicprocess. For exam-ple, in decomposing the rational functionR(x) =x+ 1(x 2)(x2+ 4)we set this up asfollows:x+ 1(x 2)(x2+ 4)=ax 2+bx+cx2+ this juncture, there are a number of approaches. One is to multiply through, clear-ing all denominators and equating coefficients in the resulting polynomial equation:x+ 1 =a(x2+ 4) + (bx+c)(x 2).

5 This quickly yieldsa+b= 0, 2b+c= 1,4a 2c= 1,from which we conclude thata= 3/8, b= 3/8, andc= 1 compute the indefinite integral R(x)dx, we need to be able to compute integralsof the form a(x )ndxand bx+c(x2+ x+ ) of the first type above are simple; a substitutionu=x will serve to finishthe job. Those of the second type can, via completing the square, be reduced tointegrals of the formbx+c(x2+a2)mdx. This involves a sum of two integrals: those of theform bx(x2+a2)mdxcan be computed via the substitutionu=x2+a2; those of theform c(x2+a2)mdxcan be handled by the appropriate trigonometric substitution(viz.,x=atan ).From the above work, we may now finish our example. x+ 1(x 2)(x2+ 4)dx=38 dxx 2 18 3x 2x2+ 4dx=38ln|x 2| 316ln(x2+ 4) +18tan 1(x2)+ Problems:1.

6 5x 3x2 2x 3dx2. 6x+ 7(x+ 2)2dx3. 2x3 4x2 x 3x2 x2 3dx4. dxx(x2+ 1)5. (1x2+ 1 1x2 2x+ 5)dx6. x3+ 2x2+ 2(x2+ 1) tan12 substitutionBasic Idea:This technique is particularly useful in computing definite integrals hav-ing integrands of the form1a+bcos or1a+bsin . If we lett= tan12 , then usingthe double-angle identity forthe tangent:tan 2A=2 tanA1 tan2A,we obtain immediately thattan =2t1 t2. 1 t22tFrom the picture depicted to the right, we obtain, therefore, thatsin =2t1 +t2and thatcos =1 t21 + We use the above to compute /2043 + 5 sin d .With the substitutiont= tan12 , we havedtd =12sec212 =1 +t22. From this it followsthatd =2dt1 +t2; we now proceed as follows: /2043 + 5 sin d t= tan12 = 1043 + 10t/(1 +t2) 21 +t2dt= 1083t2+ 10t+ 3dt= 10(33t+ 1 1t+ 3)dt=ln(3t+ 1) ln(t+ 3) 10=ln 3 Practice Problems:11.

7 /2031 + sin d 2. 2 /3035 + 4 cos d 3. /2 /234 + 5 cos d 4. /2053 sin + 4 cos d Equations Variables Idea:The IB syllabus for Calculus (Core Topic 7) contains a component relatingto a special class of differential equations, namely those having the variables separa-ble. Specifically, this relates to those differential equationsdydx=f(x,y), where thefunctionf(x,y)can be written in the formf(x,y) =g(x)h(y), for suitable functionsgandh. Such a differential equation can, in principle, yield an implicit solution foryvia separating the variables and integrating:dydx=g(x)h(y) dyh(y)=g(x)dx dyh(y)= g(x) that the integrations can be performed (which is a significant assumption!)we arrive at an equation of the typeH(y) =G(x), which definesyimplicitly as afunction (and the example above) have been lifted from Sadler and Thorning, pp 500 501:EXAMPLE1.

8 Consider the differential equationdydx= 3x2y, subject to the initialconditiony(0) = 2. We proceed as above:dydx= 3x2y dyy= 3x2dx dyy= 3x2dx ln|y|= x3+ above can be rendered more explicit by exponentiating both sides and settingK=eC(an arbitrary constant); the result isy=Ke x3. Finally, use the initial condi-tiony(0) = 2:2 =Ke0=K, and so the resulting solution isy= 2e This time, we consider the so-calledlogistic differential equationdydx=ay(1 y),wherea >0is a constant,y(0) =. separating the variables, we obtain dyy(1 y)= , using the partial fraction decomposition1y(1 y)=1y+11 y, we obtain (1y+11 y)dy= adxfrom which it follows thatln|y| ln|1 y|=ax+C y1 y= foryin terms ofxis fairly easily done; the result isy=Keax1 +Keax=11 +Be ax,whereB=K 1, again, an arbitrary conclude with a few words of terminology.

9 What we have considered above areusually calledordinary differential equations, typically abbreviated ODE. These areto be distinguished frompartial differential equations, which, as you can guess, in-volve partial derivatives and are typically much , the arbitrary constantwhich arises in the integration of an ODE is typically solved via the specification of2 One of the Millennium Problems is to help the mathematical community arrive at a better understanding of the Navier-Stokesequations, which are expressible through partial differential initial condition, often expressed in the formy(0) =y0. If both the differentialequation and the initial condition are expressed, say by writingdydx=f(x,y), y(0) =y0,we call the above aninitial value problem, or Problems:Solve the following IVPs.

10 (Unless it is convenient to do so, donot attempt to write the solutionyexplicitlyas a function ofx.) , y(0) = , y(0) = 2x(y+ 3), y(0) = +yx2 1, y(0) = 2.