The Substitution Rule
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1 The Sbstittion Rle Kiryl Tsishchanka THEOREM The Fndamental Theorem Of Calcls, Part II): If f is continos on [a,b], then where F is any antiderivative of f, that is F f. b a ] b fx)dx Fb) Fa) Fx) a NOTATION: To denote the set of all antiderivatives of f on an open) interval I we se the indefinite integral notation: fx)dx Fx)+C Table Of Indefinite Integrals cfx)dx c fx)dx [fx)+gx)]dx fx)dx+ x n dx xn+ +C n ) dx ln x +C n+ x e x dx e x +C a x dx ax lna +C sinxdx cosx+c cosxdx sinx+c sec xdx tanx+c csc xdx cotx+c secxtanxdx secx+c cscxcotxdx cscx+c +xdx arctanx+c dx arcsinx+c x gx)dx EXAMPLES:. x dx [PR with n ] x+ x +C + +C... 5 dx x x xdx x /5dx x x / dx x x x x / x +/ dx dx dx x x / x/ x /5 dx [PR with n /5] x /5+ /5+ +C 5 x/5 +C x +/ }{{} x / dx [PR with n /] x /+ /+ +C 7 x7/ +C x +/ / }{{} x 5/6 dx [PR with n 5/6] x 5/6+ 5/6+ +C 6 x/6 +C
2 Kiryl Tsishchanka 5. Table Of Indefinite Integrals cfx)dx c fx)dx [fx)+gx)]dx fx)dx+ x n dx xn+ +C n ) dx ln x +C n+ x e x dx e x +C a x dx ax lna +C sinxdx cosx+c cosxdx sinx+c sec xdx tanx+c csc xdx cotx+c secxtanxdx secx+c cscxcotxdx cscx+c +xdx arctanx+c dx arcsinx+c x 5 x+7xsinx 9x 9 5 x dx 9 x / dx+ ) dx 9x 5x/ 9x + 7xsinx 9x 7 9 sinxdx 9 x dx ln x 5 9 ) dx gx)dx 9 x 5 9 x / + 7 ) 9 sinx dx x / dx+ 7 sin xdx 9 x /+ /+ 7 9 cosx+c ln x 9 x/ 7 9 cosx+c x +x+ x +x dx x + x + + ) xx +) dx + ) dx x+ ) dx x + x + x + x + [ ] x ) ) +arctanx +arctan +arctan +arctan + π x+x ) dx x +x +x ) dx x+x +x 5 )dx x +x + x6 6 +C x + x + 6 x6 +C 8. x+x ) dx x +x +x +x 6) dx x+x +x 5 +x 7 )dx 9. x +x +x6 6 + x8 8 +C x + x + x6 + 8 x8 +C x+x ) 5 dx???
3 ANSWER : x+x ) 5 dx Kiryl Tsishchanka x +5x +75x x x 9 +96x x x x 86 +5x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x +5x x x +896x +96x +99x 8 +85x 6 +75x +5x )/+C ANSWER : x+x ) 5 dx +x ) 5 +C becase ) ) [ cf) cf +x ) 5 +C +x ) 5 +C C ] +x ) 5) [ n ) n n ] 5+x ) 5 +x ) 5+x ) 5 x x+x ) 5 THEOREM The Sbstittion Rle): If gx) is a differentiable fnction whose range is an interval I and f is continos on I, then fgx))g x)dx f)d SOLUTION: x+x ) 5 dx +x d+x ) d xdx d xdx d 5 d 5 d 5 5 +C +x ) 5 +C 5
4 Kiryl Tsishchanka cf)d c f)d n d n+ +C n ) n+ e d e +C sind cos+c sec d tan+c sectand sec+c +d arctan+c Table Of Indefinite Integrals [f)+g)]d f)d+ g)d d ln +C a d a lna +C cosd sin+c csc d cot+c csccotd csc+c d arcsin+c. 5x d 5x ) d x sin 5x )dx 5x dx d x dx 5 d 5 cos)+c 5 cos 5x )+C sin ) d 5 5 sin d. x+ x +x dx x +x x+)dx x +x dx +x ) d d x+)dx d / d /+ /+ +C x +x ) /+ +C x +x ) / +C /+. lnx x dx lnx x dx lnx dlnx) d dx d x d +C ln x +C. xe x + dx
5 Kiryl Tsishchanka cf)d c f)d n d n+ +C n ) n+ e d e +C sind cos+c sec d tan+c sectand sec+c +d arctan+c Table Of Indefinite Integrals [f)+g)]d f)d+ g)d d ln +C a d a lna +C cosd sin+c csc d cot+c csccotd csc+c d arcsin+c. xe x + dx x + dx +) d xdx d xdx d e d e d e +C ex + +C. x x+dx x+ dx+) d ) d / / )d dx d /+ /+ /+ /+ +C 5 x+)5/ x+)/ +C 5. x+5 x+ dx x+ dx+) d dx d dx d +5 ) d + d +7 d 7 d+ d d+ 7 d + 7 ln +C x+)+ 7 ln x+ +C 6. a) x+ x +x+ dx b) x+ x +x+ dx 5
6 Kiryl Tsishchanka Table Of Indefinite Integrals cf)d c f)d [f)+g)]d f)d+ g)d n d n+ +C n ) d ln +C n+ e d e +C a d a lna +C sind cos+c cosd sin+c sec d tan+c csc d cot+c sectand sec+c csccotd csc+c +d arctan+c d arcsin+c 6a). 6b) x+ x +x+ dx x+ x +x+ dx + + d x +x+ dx +x+) d d ln +C lnx +x+)+c x+)dx d x+ x+ ) x+ dx + d x+ ) d + + d dx d ) d + ln + ) + arctan +C + d+ lnx +x+)+ arctan x+ +C For more details, see Appendix I) dx x dx x)+x) x + ln x + ln +x +C ln +x x +x dx e x dx + x )dx + d + x +x ) dx x dx+ +x dx +C For more details, see Appendix II) x/ dx dx/ ) d ) dx + x dx/ d dx d + d arctan+c arctan x +C 6
7 Kiryl Tsishchanka cf)d c f)d n d n+ +C n ) n+ e d e +C sind cos+c sec d tan+c sectand sec+c a + d a arctan a +C Table Of Indefinite Integrals [f)+g)]d f)d+ g)d d ln +C a d a lna +C cosd sin+c csc d cot+c csccotd csc+c a d arcsin a +C 9. e x dx x d x) d dx d dx d e d) e d e +C e x +C. sin 5x)dx 5x d 5x) d 5dx d dx 5 d sin ) d 5 5 sin d 5 cos)+c 5 cos 5x)+C. sinx)dx 7
8 Kiryl Tsishchanka sinx)dx sinx sinxcosxdx dsinx) d cosxdx d d x dx) d sinx)dx dx d dx d sin ) d +C sin x+c sind cos)+c cosx)+c THEOREM The Sbstittion Rle for Definite Integrals): If g is continos on [a,b] and f is continos on the range of gx), then b fgx))g x)dx gb) f)d a ga). Find π/ sinx)dx. INCORRECT!!!: π/ sinx)dx π/ sinxcosxdx sinx dsinx) d cosxdx d π/ d ] π/ ] π/ π ) sin x sin sin ) METHOD : sinx)dx sinxcosxdx sinx dsinx) d cosxdx d d +C sin x+c Therefore π/ sinx)dx sin x ] π/ π ) ) sin sin. METHOD : π/ sinx)dx π/ sinxcosxdx sinx dsinx) d cosxdx d sinπ/) sin ] ) d / 8
9 Kiryl Tsishchanka. x +8 x x +8) dx +8) d dx x dx d x dx d +8 ) +8 7 d d [ ] 7 9 7, x x + dx x x+) dx x + dx +) d xdx d xdx d x ) xdx x+ d x+) d dx d x dx d x x x d x) d dx d dx )d + + / d 9+ d + 9 / d d [ ) ] d [ ] d [ 5 5/ / x+ x x ) dx+) d dx x+ dx d dx d + + ] / / )d [ / / ] 9 [ ] + [ ] d [ 5 ] 5 ) / d 9 / / )d 8. θ+cos θ ) dθ 6 θ 6 ) θ 6 θ d d 6 dθ 6d /6 /6 [ / / / / ] 9 6+cos)6d 6 /. d+6 / cos d [ 6 ] / +6[sin] /
10 Kiryl Tsishchanka To find Appendix I x+ dx, we first rewrite the denominator as x +x+ x +x+ x +x + x +x + ) ) + x+ ) ) + x+ ) + We have x+ x+ x +x+ dx x+ ) x+ dx + d x+ ) d + + d dx d + + d + d+ + d + d+ Note that and hence + v d + d + ) dv d dv d dv +a d Therefore x+ x +x+ dx a a + )d a a + + d+ x+ ) + ) d + v dv ln v +C ln + ) +C a v d dv d a) a) + d dv a a d adv v + dv a arctanv +C a arctan a +C ) d arctan )+C + v + adv ) d ln + ) + arctan )+C lnx +x+)+ arctan x+ +C
11 Kiryl Tsishchanka To find Indeed, Appendix II dx x, we first note that x x)+x) x + ) +x x + ) +x +x x)+x) + ) x x)+x) +x+ x x)+x) Therefore We have dx x x + ) dx +x dx d x dx+ x)+x) x)+x) +x dx x x dx d x) d dx d d ln +C ln x +C and +x dx +x d+x) d dx d d ln +C ln +x +C It follows that dx x x dx+ +x dx ln x + ln +x +C ln x +ln +x )+C ln +x x +C
12 Kiryl Tsishchanka EXAMPLE: Find x+ a) dx x Soltion: a) We have b) Appendix III x + x dx x+ dx x x + )dx x x +x / ) dx ] [x+ x /+ /+ ] [x+ x/ / [ x+ x ] + ) + ) b) We have x + x dx + x x x ) ) d [ ] [ +ln [ +ln) dx d ) dx )d + + )d + d + ) d + ) d + We can apply the -sbstittion in a bit different way: + x x d+ x) d x + x dx dx d x dx xd dx )d ) )] +ln +ln )] [ ln ] ln + + )d [by the above] ln REMARK: Problem b) was given in Fall Calcls II, qiz ). Nobody solved this problem correctly.
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