Prob9sol.pdf - MAT 21B(Winter 2019 Problem session Sections...

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MAT 21B (Winter 2019)Problem session: Sections 8.4, 8.5SOLUTIONS – please report errors/typosEvaluate the following integrals:1.ż?a2´x2dxLetxasinθ. Thenż?a2´x2dxżaa2´a2sin2θ dpasinθq “ża2cos2θ dθa22żp1`cos 2θqa22θ`a24sin 2θ`Ca22θ`a22sinθcosθ`Ca2´1xa`x2?2sina2´x2`C .2.żx?4´x2dxLetx2 sinθ. Thenżx?4´x2dxż2 sinθa4´4 sin2θ dp2 sinθq “8żsinθcos2θ dθ“ ´83cos3θ`C“ ´83´1´x24¯3{2`C“ ´13p4´x2q3{2`C .Alternatively:this is easy to do without a trig substitution!Just setu4´x2,du“ ´2x dxto getżx?4´x2dx“ ´12ż?u du“ ´13u3{2`C“ ´13p4´x2q3{2`C .3.ż1?e2t`1dtAgain, there are multiple ways to approach this. To set it up for a trig substitution, letxet(ortlnx) anddt1xdx. Thenż1?e2t`1dtż1x?x2`1dxxtanθż1tanθ?tan2θ`1dptanθq “ż1tanθsecθsec2θ dθżcscθ dθ“ ´ln|cscθ`cotθ| `C“ ´lnˇˇˇ?1`x2x`1xˇˇˇ`Ct´ln|?e2t`1`1| `C .Alternatively: try a substitutionu?e2t`1 (ore2tu2´1). Thendue2t?e2t`1dt,which we rearrange to getduu2´1dt?e2t`1. Therefore,ż1?e2t`1dtż1u2´1du .
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We can now do this by partial fractions, or with a trig substitutionusecθ.The trigsubstitution ultimately reproduces the same answer as above The partial fractions giveż1u2´1du12ż1u´1´1u`1ıdu12ln|u´1| ´12ln|u`1| `C12ln|?e2t`1´1| ´12ln|?e2t`1`1| `C .After some simplification, this turns out to be the same answer as above!4.żx29`4x2dxOne option is to do a trig substitution right away. Let’s factor out a 4 in the denominatorand setx32tanθto getżx29`4x2dx14żx29{4`x2dx14ż9{4 tan2θ9{4 sec2θp3{2qsec2θ dθ38żtan2θ dθ38żpsec2θ´1q38tanθ´38θ`C14x´38tan´12x3`C .Alternatively, notice that the numerator has the same degree as the denominator. So wecan simplify the rational function by dividing:żx29`4x2dxż14´9{49`4x2ıdx14ż1´11` p4{9qx2ıdx14x´38tan´12x3`C .5.żx?1`x4dx[Hint: substituteux2(orx?u) first.]Settingx?u

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