1、Integration by substitution and by parts in definite integrals1Integration by Substitutions for definite integrals2,()baf x dx(),()F xCf x dx When we want to evaluate the value of a definite integral if we can find the corresponding indefinite integral then we can use the Newton-Leibniz formula to o
2、btain the value immediately.()()()()()dbbbaaaf x dxF bF aF xF x Integration by Substitutions for definite integrals3(Integration by substitution for definite integrals)Suppose that the function x=(t)is continuously differentiable on the interval,()=a,()=b,and that f is continuous in the range of ,I.
3、Then()()()baf x dxftt dt Integration by Substitutions for definite integrals4Proof()()()baf x dxF bF a ()()(),dFtfttdt ()()()ftt dtFt Let F be an antiderivative of f on the interval I,thenSince we haveTherefore,the formula holds.()()FF ()()F bF aIntegration by Substitutions for definite integrals5si
4、n,xt cos.dxtdt 0t 0;x 2t 1;x 1222001cosx dxtdt 1201.x dx Evaluate Let Then when when thus by the formula,we haveSolution2011sin222tt.4 Integration by Substitutions for definite integrals60202sinsin22nnxdxt dt 2200sincosnnxdxxdx Prove (n is any positive integer).Proof02cosntdt 20cosntdt 20cos.nxdx In
5、tegration by Substitutions for definite integrals7 Find 2001sinlimd.xxxtttx Solution Let,xtu then,utx dd.utx 0t 0;u2.uxtx Then0sindxxttt 20sinxu duuxx 20sindxuuu 2022000sind1sinlimdlimxxxxuuxtutxtx So 00220sin2lim2xxxxx 1.Integration by Parts for indefinite integrals 8When u and v are differentiable
6、 functions of x,the Product Rule for differentiation tell us that()ddvduuvuvdxdxdxIntegrating both sides with respect to x and rearranging leads to the integral equation()dvdduudxuvdxvdxdxdxdx.duuvvdxdx Integration by Parts for indefinite integrals When u and v are differentiable functions of x on t
7、he interval ,a b.Then bbbaaaudvuvvdu 92.dxtdt 42200022xttedxetdttdext Let,then 2xt,Thus Solution 22002tttee dt 22(1).eExample Evaluate 40.xedx Integration by Parts for indefinite integrals 10 Evaluate 120arcsin.xdx Solution Letarcsin,ux,dvdx then2,1dxdux ,vx 120arcsin xdx 120arcsinxx 12201xdxx 12 6
8、1222011(1)21dxx 12 12201x31.122 bbbaaaudvuvvduIntegration by Parts for indefinite integrals bbbaaaudvuvvdu11 Evaluate 40.1 cos2xdxx Solution21cos22cos,xxQ401cos2xdxx 4202cosxdxx 40tan2xdx 401tan2xx 401tan2xdx 401lnsec82x ln2.84 Integration by Parts for indefinite integrals bbbaaaudvuvvdu12 Evaluate
9、120ln(1).(2)xdxx Solution120ln(1)(2)xdxx 101ln(1)2x dx 10ln(1)2xx 101ln(1)2dxx ln23 101121dxxx 1112xx 10ln2ln(1)ln(2)3xx 5ln2ln3.3Integration by Parts for indefinite integrals 13201)sinnnIxdx bbbaaaudvuvvdu Compute the following integrals:4222001)sin;2)sincos.nnIxdxxxdx Solution(1)20sin(cos)nxdx 222
10、0(1)(1sin)sinnnxxdx 22200(1)sin(1)sinnnnxdxnxdx 2220(1)cossinnnxxdx 2(1)(1).nnnInI 1222200sincos(1)cossinnnxxnxxdx 21(2,3,)nnnIInn Integration by Parts for indefinite integrals 144222001)sin;2)sincos.nnIxdxxxdx Solution(continued)bbbaaaudvuvvdu241132nnnnnnIIInnn 113 2,2 3nnInn 241132nnnnnnIIInnn If
11、n is odd013 1.2 2nnInn If n is even210sin1,Ixdx 200,2Idx Since134 2,ifis odd;25 3133 1,ifis even.24 2 2nnnnnnInnnnn we obtain Compute the following integrals:Integration by Parts for indefinite integrals 15424222002)sincossin(1sin)xxdxxx dx134 2,ifis odd;25 3133 1,ifis even.24 2 2nnnnnnInnnnn Comput
12、e the following integrals:4222001)sin;2)sincos.nnIxdxxxdx Solution(continued)bbbaaaudvuvvdu462200sinsinxdxxdx3 15 3 14 2 26 4 2 2.32 Quadrature Problems for elementary fundamental functions16By the previous examples,we have seen that quadratures are much more difficult than differentiations.When int
13、egrands are continuous,their integral must exist,but their computation sometimes requires skill,and sometimes may nor even be expressible by elementary functions.For instance,the integrals:24sin,1xxdxe dxdxxx seems very simple,and the integrands are all continuous.All of these integrals exist,but we can not express them in terms of elementary functions.In general,we have known that for any rational function and any rational trigonometric function,their integrals can be expressed by elementary functions.
