Integration. September 28, 2017

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1 Integrtion September 8, 7 Introduction We hve lerned in previous chpter on how to do the differentition. It is conventionl in mthemtics tht we re supposed to lern bout the integrtion s well. As you my hve lredy know, integrtion is the inverse of differentition, or sometimes clled ntiderivtive. In fct, by relising the reltionship between these two opertions, hs lid out the foundtion of wht we now clled Clculus which ws invented by Newton nd Leibniz independently in 7th century. Our min gol for this chpter is to lern bout the integrtion of the trnscendentl functions. First we will strt on reviewing the integrtion of simple integrnd(the eqution tht we wnt to integrte, the indefinite nd the definite integrl. Next we will lern the three techniques of integrtion, nd then we will pply these techniques on integrting trnscendentl functions ccordingly. Review of Integrtion. Bsic Integrl Formul Exmple : Evlute the following indefinite integrls. x x x 3 x

2 sin x cos x e x x x 8 x 4 5 x 3 x 5 4 sec x cosec x ( + tn x sec x sin x cos x sin x cos x. Integrls with Sum, Difference nd Fctor Exmple : Evlute the following indefinite integrls. (3x + 8. ( 4 sech x tnh x

3 (x 4 + x 6 (3x x 4 x 3 (4 + tn x (sin x cos x (x + (3x 8. + e x sec x sec x.3 Definite Integrl Exmple 3: Evlute the following integrls.. 3. (3x + 7 (x + (x + 3 x 3 + x 3 3 Techniques of Integrtion 3. Integrtion by Substitution Method of substitution is strightforwrd from chin rule, nd this method is wht you should consider first before other methods. We usully choose the substitution for the inside function nd then mnipulte some lgebr for wht s left on the outside function. Exmple 4: Evlute the following integrl by using the proper substitution. (4x +. c x + b 3

4 e 4x e 3x+5 (x 9 5 sin 5x cos(x + 3 tn(3 x 9. x + 3 x + 3x +. x x 4 x + 6. x 3 + 4x. x + cos x 3x + 6 sin x 3. tn x sin x cos x sec x 4x(x 3 6 sin x cos 4 x e x (e x 3 9. ln x x 4

5 x ln x x sec (x + 3 tn(x + 3 xe x x(x + 5 x 3x x x x 3 cos x 4 e x sin e x e tn x sec x 5x cos(5 x + x + e x x x 4 sin (6x + 7 Sometimes the substitution is given to us. Hve in mind tht the substitution will help us to simplify the integrnd, not mking it more difficult to integrte, which is the sole reson on why we re lerning to find the pproprite substitution in using this technique. Exmple 5 Evlute the following integrls by using the given substitution. 3x 3x with substitution u = 3x. 5

6 . 4x with substitution x = sin θ. x 3. 4 x with substitution x = sin θ. 3. Integrtion by Prts Integrtion by prts is powerful method to evlute integrls when the integrnd is in the form of product of lgebric nd trnscendentl, such s x ln x, xe x, e x cos x. Generl formul for integrtion by prts is u dv = uv v du Tips If the integrl is in the form of x n ln x, tke u = ln x nd dv = x n. If the integrl is in the form of e x sin bx or e x cos bx, tke u = sin bx nd dv = e x. Or, we cn use the substitution the other wy round by tking u = e x nd dv = sin x. If the integrl is in the form of x n e x, x n sin x or use the substitution u = x n nd dv = e x, sin x, cos x. x n cos x, Exmple 6 Evlute the following integrl using the following integrls. x ln x. ln x 3. xe x xe x x sin x 6

7 6. 7. π x cos x x cos x π x sin x e x cos x x ln x 3.3 Tbulr Method When integrting by prts hve mny repeted integrtions nd differentitions, tbulr method is very useful trick since it s mking the integrtion by prt procedure more net. Exmple 7 Evlute the integrl using the tbulr method. x 3 e x.. x sin 3x. 3. x sec x. 4. x (x e 3x cos x 6. cos 5x sin 4x 3.4 Integrtion by Prtil Frction By using this technique, we rewrite the integrl which initlly in product nd quotient form, in terms of frction. In prctice the integrl is in the form of polynomil of rtionl functions, where the denomintor usully cn be fctorised. Though pretty much most of polynomil of degree cn be fctorised, ber in mind tht if the 7

8 fctor on the numertor results to complicted lgebr, you re dvised to use other method ie substitution or inverse trigonometric nd inverse hyperbolic. If the numertor hs the sme order of degree or more with the denomintor, it becomes n improper frction. For this prticulr cse, you re required to use long division of polynomils. Exmple 8 Evlute the integrl by using prtil frctions. 3x + x + 3x +. x 3 3x + x x + 7 (x x + x + 6 x 3 + x + x x x + x 3 x x x + 3x + x 5 5x (x + 4 Integrls of Hyperbolic Functions Integrls of hyperbolic functions re similr to the integrls of trigonometric functions. You my refer to the formul to find the ntiderivtive of the hyperbolic functions. As for the more complicted integrls, we cn use the three methods described erlier. Exmple. Evlute the following integrls sinh x b 5 cosh 3x c sinh x 8

9 d e f g cosh x 3 sech 4x cosh (5x + cosh x + 3 sinh x. By using the definitions show tht Hence show tht cosh x + sinh x = cosh x sinh x. cosh x + sinh x = ( e. 3. By using the tbulr method, evlute the following integrl x 3 cosh x b x sinh 3x c e 3x cosh x d sin x sinh 3x 5 Integrtion Involving Inverse Hyperbolic Functions nd Inverse Trigonometric Functions When integrtion involving inverse hyperbolic nd inverse trigonometric functions re in the form of x sinh - x, x cosh - x, tn - x, they cn be integrted using integrtion by prt(ibp sin - x, u dv = uv v du. It is solvble if we choose the inverse functions s u in the IBP formul, since the inverse function cnnot be integrted directly, nd cn only undergo differentition. Another form is when the integrnd is in the form of frction. Usully when we re given n integrl in frction form, we first try the method of substitution. If it is 9

10 not working, the we try the method of prtil frction. It is not necessrily in tht order, it cn be interchnge. However if both methods re not pplicble, then it must hve involved inverse hyperbolic or inverse trigonometric. inverse hyperbolic/trigonometric is in the form of Ax + B or Ax + B or Generlly the integrtion tht will result to x Ax + B. (5. Another form is Ax + Bx + C or Ax + Bx + C or x Ax + Bx + C, (5. tht requires completing the squre to get the form in Eqution 5.. Here, x is ny function of x. Once we hve the form s in Eqution 5., we just need to mtch it with the formul given by using the method of substitution to get the ntiderivtive tht involve the inverse trnscendentl function. It would be useful if we fmilirize ourselves with the technique of completing the squre tht will be used frequently in trnsforming the integrnd. Completing the squre is trnsforming the qudrtic eqution into x + bx + c =, (x + d + e =, where d = b b nd e = c 4. Prctice : Rewrite this expression by completing the squre. x + 4x +. x 6x 3. x 8x x + 7x + 3 Another pproch is to mke trigonometric substitution, which is originlly how the formul mterilized to the generl form s we cn use now. The form nd

11 respectively the substitution re given s : if x, substitute x = sin θ or x = cos θ (5.3 if + x, substitute x = sinh θ or x = tn θ (5.4 if x, substitute x = cosh θ or x = sec θ, (5.5 where is constnt. 5. Integrtion Involving Inverse Trigonometric Functions In previous chpter we hve lerned how to do the differentition of the inverse function using the formul. Hence to find the integrl, the ide is the sme. Wht we need to do is to mnipulte the integrnd(provided tht we re sure tht the integrl will result to inverse trigonometric or inverse hyperbolic, so tht it mtches the formul. Recll tht the differentition of inverse trigonometric is given s d [sin- x] = x d [cos- x] = x d [tn- x] = + x d [cot- x] = + x d [sec- x] = x x d [cosec- x] = x x.

12 Therefore the integrl formul is the ntiderivtive of those nd given s = x sin- x + C, x < = x cos- x + C, x < + x = tn- x + C + x = cot- x + C x x = sec- x + C, x > x x = cosec- x + C, x > Generlly, the integrl for the inverse trigonometric will look like this x, nd t first glnce, we know it looks like the differentition eqution of we let x = u, then = du. Hence x = du ( x u = sin- + C. d [sin- x]. If The sme procedure pplied on the pproprite integrnd similr to the form of differentition of tn - x nd sec - x. The generl formul for this type of integrnds re given s ( x x = sin- + C, x < + x = ( x tn- + C x x = ( x sec- + C, x >, where the integrtion formul tht ssocites with cos - x, cot - x nd cosec - x re just the negtive terms of the bove integrnds respectively. Exmple Evlute the following integrls.. 5 x. 9 + x

13 x x 4 9x 4 + 9x x 4x x x x + 4x + 5 x + x + x / x x x + (x x x 3 sin - x cos - x x tn - x 5. Integrtion Involving Inverse Hyperbolic Functions The procedure of solving the integrtion involving inverse hyperbolic functions is similr to the procedure of solving the integrting involving inverse trigonometric 3

14 functions. The generl formul for the integrnd tht will result to the inverse hyperbolic functions re given s ( x + x = sinh- x + C, > ( x x = cosh- x + C, x > x = ( x tnh- + C, if x < x = ( x coth- + C, if x > ( x x x = sech- + C, if < x < ( x x + x = cosech- + C, if < x < Exmple Evlute the following integrl.. x + x e x. 6 e x x x 6, x > 4 4 x, x < x 4x x x 4x + 5 4

15 . 5 3 x x.. (x + x + x + x x tnh - x 6 Conclusion In this topic, you re required to know how to integrte the integrls tht involve trigonometric, inverse trigonometric, hyperbolic nd inverse hyperbolic functions. Wht you should know first is to mster ll the techniques of integrtion which re Integrtion by Substitution Integrtion by Prts( nd Tbulr Method Integrtion by Prtil Frctions Once you hve fmilir yourselves with ll these techniques, the next step is to recognised the pttern of which technique you should use given n integrl. There is no trick or shortcut to it. It is dvised for you to do s mny integrtion prctises s you cn, then you will be ble to recognised the pttern. However it is suggested tht (but don t tke the following s strict rule tht you hve to follow for you to lwys try the method of substitution first. If it s not working ie the resulting substitution is not simple lgebric eqution, then you cn try prtil frction (if the integrl is in frction, nd followed by integrtion by prts. If you were given n integrtion in the form of frction with the denomintor is in squre root form, try the method of substitution first, if it isn t working, see if the integrtion of inverse trigonometric or hyperbolic works. If you were given n integrtion in form of frction of polynomil, check the order of the power. If the order of the numertor is the sme or greter thn the denomintor, then use long division first. 5