7.5 Integrals Involving Inverse Trig Functions

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1 . integrls involving inverse trig functions. Integrls Involving Inverse Trig Functions Aside from the Museum Problem nd its sporting vritions introduced in the previous section, the primry use of the inverse trigonometric functions in clculus involves their role s ntiderivtives of rtionl nd lgebric functions. Ech of the six differentition ptterns from the previous section provides us with n integrl formul, but they give rise to only three essentilly different ptterns: = rcsinx + C x = rctnx + C + x x = rcsecx + C x Most of the relted ntiderivtive ptterns you will need in prctice rise from vritions of these bsic ones. Typiclly, you need to trnsform n integrnd so tht it exctly mtches one of the bsic ptterns. Exmple. Evlute 6 + x. Solution. We cn trnsform this integrnd into the rctngent pttern by fctoring 6 from the denomintor 6 + x = 6 + x 6 = 6 + nd then using the substitution u = x du = du = : 6 + u du = 6 + u du = rctnu + C Replcing u with x we get rctn + C s our finl nswer. Prctice. Evlute + 9x nd. x The integrnds tht rise most often contin ptterns with the forms x, + x nd x, where is some positive constnt, so it is worthwhile to develop generl integrl ptterns for these forms, list them in Appendix I, nd refer to them when necessry: x = rcsin + C + x = rctn + C x x = rcsec + C Vlid for: < x < Vlid for: < x < Vlid for: x > Why re the derivtive ptterns for rccos, rccot nd rccsc of little use to us when finding ntiderivtives of lgebric functions? Vlid for: < x < Vlid for: < x < Vlid for: x >

2 6 trnscendentl functions You cn rrive t ech of these generl formuls by fctoring the out of the denomintor nd mking suitble chnge of vrible s in Exmple, with in plce of. You cn then check the result by differentiting. The rctn pttern is, by fr, the most common. The rcsin pttern ppers occsionlly, nd the rcsec pttern only rrely. Exmple. Develop the generl formul for from the x known formul for. Assume tht >. x Solution. Using Exmple s guide, fctor out of the denomintor: x = = x Now substitute u = x du = du = to get: du = u nd replce u with x to get rcsin u du = rcsinu + C + C, the desired result. Prctice. Verify tht the derivtive of rctn is Exmple. Evlute nd x + x. + x. Solution. The constnt needn t be n integer, so tke = = : x = rcsin + C x using the pttern from Exmple, while the generl rctn pttern yields: [ ] x + x = rctn = ] [rctn rctn or bout.8. The esiest wy to integrte certin rtionl functions is to split the originl integrnd into two pieces. Exmple. Evlute + + x.

3 . integrls involving inverse trig functions Solution. The integrnd splits nicely into the sum of two other functions tht you cn integrte more esily: + + x = + x + + x In the first integrl, use the subsitution u = + x du = x du = x : + x = 6 u du = ln u + C = ln + x + C Menwhile, the second integrl mtches the generl rctngent pttern with = : + x = Combining these two results yields: s our finl nswer. + x = rctn + C + + x = ln + x + rctn + C Why cn we remove the bsolute vlue signs in the lst step? The two constnts C nd C dd up to nother rbitrry constnt, which we cn simply cll C. The ntiderivtive of liner function divided by n irreducible qudrtic polynomil will typiclly result in the sum of logrithm nd n rctngent. x + Prctice. Evlute x +.. Problems In Problems, evlute the integrl x. 9 y x +. x x x 6. x x 8. x +. e x + e x. e + x x + 9 x x + x + [lnx] cosθ dθ. 9 sin θ x 9 + x x 8. + x. + x. x. 8x 6 + x x 9 x 9 x 9x x x + x

4 8 trnscendentl functions In Problems 8, solve the initil vlue problem. = y, y = e x = y + x, y = = y 9 + x, y = 6 x = y, y = In Problems 9, evlute the integrl by splitting the integrnd into two simpler functions. 8x + x 9. x + 9. x + x +. x +. x + x + 6 Problems illustrte how we cn sometimes decompose difficult integrl into simpler ones. Hints: For, complete the squre in the denomintor; for, let u = denomintor; for, write x + = x x + + x + x + + x + x + x + + x + x x + x + + x + x + x + x + x + x + x x +. Prctice Answers. For the first integrl, write: + 9x = + x nd substitute u = x du = du = to get: + u du = + u du = rctnu + C = rctnx + C For the second integrl, fctor out to get: = = x x nd then substitute u = x du = du = : = x du = rcsinu + K u Replcing u with x yields: = rcsin + K x. Using the Chin Rule: [ ] D rctn = + x = [ + ] = + x

5 . integrls involving inverse trig functions 9. Split the integrnd into two pieces: x + x + = x x + + x + For the first integrl, let u = x + du = x du = x : x x + = u du = ln u + C = ln x + + C The second integrl mtches the rctngent pttern with = = : x + = x = rctn + C + x Combining these results yields: x + x + [ = ln x + ] + x rctn + C