Trignometric Substitution Trigonometric substitution refers simply to substitutions of the form x sinu or x tnu or x secu It is generlly used in conjunction with the trignometric identities to sin θ+cos θ 1 nd 1+tn θ sec θ eliminte x from n integrnd by substituting x sinu to give x sin u cos u cosu or to eliminte +x from n integrnd by substituting x tnu to give +x + tn u sec u secu or to eliminte x from n integrnd by substituting x secu to give x sec u tn u tnu. When we hve used substitutions before, we usully gve the new integrtion vrible, u, s function of the old integrtion vrible x. Here we re giving the old integrtion vrible, x, in terms of the new integrtion vrible u. We my do so, s long s we my invert to get u s function of x. For exmple, with x sinu, we my tke u rcsin x. This is good time for you to review the definitions of rcsinθ, rctnθ nd rcsecθ. See the notes Inverse Functions. Exmple 1 ( r x ) Let s find the re of the shded region in the sketch below. y x +y r r x We ll set up the integrl using verticl strips. The strip in the figure hs width nd height r x. So the re is r x. To evlute the integrl we substitute x rsinu rcosu du becuse then we will be ble to use r x r r sin u r ( 1 sin u ) r cos u c Joel Feldmn. 15. All rights reserved. 1 Februry, 15
to eliminte the squre root from the integrnd. Let s think bout the limits of integrtion. Our integrl hs x running from x to x r. The vlue of u tht corresponds to x r is u π / (which solves x r rsinu, i.e. which solves sinu 1) nd the vlue of u tht corresponds to x is u rcsin /r (which solves x rsinu. i.e. which solves sinu ). As u runs from u rcsin r /r to u π, x rsinu runs from x to x r covering exctly the domin of integrtion. So we ll mke the domin of integrtion, in the u integrl, rcsin /r u π. We re now redy to do the integrl. r x π/ rcsin /r r r sin u rcosu du with x rsinu, rcosu du π/ rcsin /r π/ rcsin /r r cos u rcosu du r cos u du Becreful bouttkingthesqurerootinthelststep. Becuse r x denotesthepositive squre root of x, r cos u denotes the positive squre root of r cos u. Fortuntely, the domin of integrtion is contined in u π nd cosu there. So rcosu relly is thepositive squre rootofr cos uinour integrl. If ourdominofintegrtionhdcontined u s between π nd π, for exmple, we would hve needed to write r cos u r cosu. Now bck to evluting the integrl. r x π/ r r rcsin /r π/ rcsin /r r cos u du u+ sin(u) ] 1+cos(u) du since cos u 1+cos(u) ] π/ rcsin /r r π rcsin r sin(rcsin /r) To simplify sin(rcsin /r), let s write rcsin /r θ. Then θ is the ngle in the tringle on the right below. By the double ngle formul for sin(θ) ] So our finl nswer is sin(θ) sinθ cosθ r r r r θ r Are r x πr 4 r rcsin r 1 r (1) c Joel Feldmn. 15. All rights reserved. Februry, 15
This is reltively complicted formul, but we cn mke some resonbleness checks, by looking t specil vlues of. If the shded region, in the figure t the beginning of this exmple, is exctly one qurter of disk of rdius r nd so hs re 1 4 πr. Subbing into (1) does indeed give 1 4 πr. At the other extreme, if r, the shded region disppers completely nd so hs re. Subbing r into (1) does indeed give, since rcsin1 π. Exmple 1 Exmple ( x r x ) The integrl x r x looks lot like the integrl we just did in Exmple 1. It cn lso be evluted using the trigonometric substitution x rsinu. But just becuse you hve now lerned how to use trig substitution doesn t men tht you should forget everything you lerned before. This integrl is much more esily evluted using the simple substitution u r x. x r x du u r ] u / with u r x, du x 1 / 1 r ] / r Exmple Exmple ( x ) 9+x This time we ll substitute x tnu sec u du becuse then we will be ble to use 9+x 9+9tn u 1+tn u sec u secu to eliminte the squre root from the integrl. Note tht, to stisfy x tnu, we cn tke u rctn x, with rctn being the stndrd rctngent tht lwys tkes vlues between π / nd + π /. So u will lwys tke vlues between π / nd + π / nd cosu will lwys be positive, so tht secu secu. So our integrl x 9+x 1 9 sec u du 9tn u secu secu tn u du with x tnu, sec u du c Joel Feldmn. 15. All rights reserved. Februry, 15
1 cosu 9 sin u 1 dy 9 y 1 9y +C 1 9sinu +C du since secu 1 cosu nd 1 tn u cos u sin u with y sinu, dy cosu du The originl integrl ws function of x, so we still hve to rewrite sinu in terms of x. Remember tht x tnu or u rctn x. So u is the ngle shown in the tringle below nd we cn red off the tringle tht x 9+x sinu 9+x x 9+x x 9+x +C 9x u Exmple Exmple 4 ( 5 x x x 1 ) This time we hve n integrl with squre root in the integrnd, but the rgument of the squre root, while qudrtic function of x, is not in one of the stndrd forms x, +x, x. The reson tht it is not in one of those forms is tht the rgument, x x, contins term, nmely, x tht is of degree one on x. So we try mnipulte it into one of the stndrd forms by completing the squre, which mens tht we try to express x x in the form (x ) +b for some constnts nd b. Observe tht if we squre out (x ) +b we get x x+ +b, which will be exctly x x if we choose nd b so tht (to give the correct coefficient of x) nd +b (to give the correct constnt term). So 1, b 4 works nd we now know tht x x (x 1) 4 Then to convert the squre root of the integrnd into stndrd form, we just mke the simple substitution y x 1. Here goes 5 x x 5 (x 1) 4 x 1 x 1 4 y 4 dy y π/ 4sec u 4 secutnu du secu To get the limits of integrtion we used tht with y x 1, dy with y secu, dy secutnu du c Joel Feldmn. 15. All rights reserved. 4 Februry, 15
the vlue of u tht corresponds to y obeys y secu cosu tht u works nd or cosu 1, so the vlue of u tht corresponds to y 4 obeys 4 y secu cosu or cosu 1, so tht u π / works. Now returning to the evlution of the integrl, we simplify nd continue. 5 x x x 1 π/ π/ π/ sec u 1 tnu du tn u du tnu u ] π / since sec u 1+tn u sec u 1 ] du since sec u 1+tn u, gin ] π/ Exmple 4 c Joel Feldmn. 15. All rights reserved. 5 Februry, 15