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Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingpx 2 a2 Integrals involvingx 2 a2Trigonometric SubstitutionTo solve integrals containing the following expressions;pppa2 x 2x 2 a2x 2 a2 ,it is sometimes useful to make the following substitutions:Expressionqa2 x 2pa2 x 2qx 2 a2Substitutionx a sin θ, π θ π or θ sin 1 x22ax a tan θ, π θ π or θ tan 1 x22ax a sec θ, 0 θ π or π θ 3π or θ sec 1 x22aIdentity1 sin2 θ cos2 θ1 tan2 θ sec2 θsec2 θ 1 tan2 θNote The calculations here are much easier if you use the substitution inreverse: x a sin θ as opposed to θ sin 1 xa .Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2a2 x 2We make the substitution x a sin θ, π2 θ π2 , dx a cos θdθ,p a2 x 2 a2 a2 sin2 θ a cos θ a cos θ (since π2 θ π2 by choice. )ExampleZx3 dx4 x2Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2a2 x 2We make the substitution x a sin θ, π2 θ π2 , dx a cos θdθ,p a2 x 2 a2 a2 sin2 θ a cos θ a cos θ (since π2 θ π2 by choice. )ExampleZIx3 dx4 x2Let x 2 sin θ, dx 2 cos θdθ,Annette Pilkingtonp 4 x 2 4 4 sin2 θ 2 cos θ.Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2a2 x 2We make the substitution x a sin θ, π2 θ π2 , dx a cos θdθ,p a2 x 2 a2 a2 sin2 θ a cos θ a cos θ (since π2 θ π2 by choice. )ExampleZIIx3 dx4 x2p 4 x 2 4 4 sin2 θ 2 cos θ.Let x 2 sin θ, dx 2 cos θdθ,RRRR x 3 dx3cos θdθ) 8 sin θ(2 8 sin3 θdθ 8 sin2 θ sin θ dθ 2cosθ2R 4 x8 (1 cos2 θ) sin θ dθ.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2a2 x 2We make the substitution x a sin θ, π2 θ π2 , dx a cos θdθ,p a2 x 2 a2 a2 sin2 θ a cos θ a cos θ (since π2 θ π2 by choice. )ExampleZIIIx3 dx4 x2p 4 x 2 4 4 sin2 θ 2 cos θ.Let x 2 sin θ, dx 2 cos θdθ,RRRR x 3 dx3cos θdθ) 8 sin θ(2 8 sin3 θdθ 8 sin2 θ sin θ dθ 2cosθ2R 4 x8 (1 cos2 θ) sin θ dθ.Let w cos θ, dw sin θ dθ,ZZZ8w 38 (1 cos2 θ) sin θ dθ 8 (1 w 2 ) dw 8 (w 2 1) dw 8w C3 8(cos θ)33 8 cos θ C 8(cos(sin 1 x2 ))33Annette Pilkington 8 cos(sin 1 x2 ) C .Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingZ px 2 a2 Integrals involvingx 2 a2a2 x 28(cos(sin 1 x2 ))3x3xdx 8 cos(sin 1 ) C .2324 xAnnette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingZI px 2 a2 Integrals involvingx 2 a2a2 x 28(cos(sin 1 x2 ))3x3xdx 8 cos(sin 1 ) C .2324 xTo get an expression for cos(sin 1 x2 ), we use an appropriate triangleFrom the triangle, we get4 x 2cos(sin 1 x2 ) 22xθ4 - x2Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingZI px 2 a2 Integrals involvingx 2 a2a2 x 28(cos(sin 1 x2 ))3x3xdx 8 cos(sin 1 ) C .2324 xTo get an expression for cos(sin 1 x2 ), we use an appropriate triangleFrom the triangle, we get4 x 2cos(sin 1 x2 ) 22xθ4 - x2“ IhenceR x3dx4 x 28 4 x 223”3 8Annette Pilkington4 x 22 C (4 x 2 )3/23Trigonometric Substitution 4 4 x2 C

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleRx2 px 2 a2 Integrals involvinga2 x 2 dx9 x 2Annette PilkingtonTrigonometric Substitutionx 2 a2

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIRx2 px 2 a2 Integrals involvingx 2 a2a2 x 2 dx9 x 2Let x 3 sin θ, dx 3 cos θdθ,Annette Pilkingtonp 9 x 2 9 9 sin2 θ 3 cos θ.Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIIRx2 x 2 a2 Integrals involvingx 2 a2a2 x 2 dx9 x 2Let x 3 sin θ, dx 3 cos θdθ,RRR 3 cos θdθ dx (9 sin2 θ)3 cos θ x2p9 x 2Annette Pilkingtonp 9 x 2 9 9 sin2 θ 3 cos θ.1dθ9 sin2 θ cot θ9 C Trigonometric Substitution cot(sin 1 x3 )9 C

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingRExampleIIIx2 x 2 a2 Integrals involvingx 2 a2a2 x 2 dx9 x 2Let x 3 sin θ, dx 3 cos θdθ,RRR 3 cos θdθ dx (9 sin2 θ)3 cos θ x2p9 x 2To get an expression forp 9 x 2 9 9 sin2 θ 3 cos θ.1dθ9 sin2 θcot(sin 1 x3 ), cot θ9 C cot(sin 1 x3 )9 Cwe use an appropriate triangleFrom the triangle, we get9 x 2 1 xcot(sin 3 ) and hencex Zdx 9 x2 C229xx 9 x3xθ9 - x2Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingRExampleIIIx2 x 2 a2 Integrals involvingx 2 a2a2 x 2 dx9 x 2Let x 3 sin θ, dx 3 cos θdθ,RRR 3 cos θdθ dx (9 sin2 θ)3 cos θ x2p9 x 2To get an expression forp 9 x 2 9 9 sin2 θ 3 cos θ.1dθ9 sin2 θcot(sin 1 x3 ), cot θ9 C cot(sin 1 x3 )9we use an appropriate triangleFrom the triangle, we get9 x 2 1 xcot(sin 3 ) and hencex Zdx 9 x2 C229xx 9 x3xθ9 - x2I CNote You can also use this method to derive what you already knowZx1 dx sin 1 C22aa xAnnette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a tan θ, π2 θ π2 , dx a sec2 θdθ, x 2 a2 a2 tan2 θ a2 a sec θ a sec θ (since π2 θ π2 by choice. )ExampleZ Annette Pilkingtondxx2 4Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a tan θ, π2 θ π2 , dx a sec2 θdθ, x 2 a2 a2 tan2 θ a2 a sec θ a sec θ (since π2 θ π2 by choice. )ExampleZI dxx2 4Let x 2 tan θ, π2 θ π2 , dx 2 sec2 θdθ, x 2 4 4 tan2 θ 4 2 sec2 θ 2 sec θ.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a tan θ, π2 θ π2 , dx a sec2 θdθ, x 2 a2 a2 tan2 θ a2 a sec θ a sec θ (since π2 θ π2 by choice. )ExampleZII dxx2 4Let x 2 tan θ, π2 θ π2 , dx 2 sec2 θdθ, x 2 4 4 tan2 θ 4 2 sec2 θ 2 sec θ.RR dxR 2 sec2 θdθ sec θdθ ln sec θ tan θ C 2 sec θ2x 4ln sec(tan 1 x2 ) tan(tan 1 x2 ) C ln sec(tan 1 x2 ) x2 CAnnette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a tan θ, π2 θ π2 , dx a sec2 θdθ, x 2 a2 a2 tan2 θ a2 a sec θ a sec θ (since π2 θ π2 by choice. )ExampleZII dxx2 4Let x 2 tan θ, π2 θ π2 , dx 2 sec2 θdθ, x 2 4 4 tan2 θ 4 2 sec2 θ 2 sec θ.RR dxR 2 sec2 θdθ sec θdθ ln sec θ tan θ C 2 sec θ2x 4Iln sec(tan 1 x2 ) tan(tan 1 x2 ) C ln sec(tan 1 x2 ) x2 CTo get an expression for sec(tan 1 x2 ), we use an appropriate triangleFrom the triangle, we getx 2 4 1 xsec(tan 2 ) 2and hence Zdxx2 4 x ln C .22x2 4x2 4xθ2Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a tan θ, π2 θ π2 , dx a sec2 θdθ, x 2 a2 a2 tan2 θ a2 a sec θ a sec θ (since π2 θ π2 by choice. )Note You can also use this substitution to get the familiarZ1x1dx tan 1 C .x 2 a2aaAnnette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square.Sometimes we can convert an integral to a form where trigonometricsubstitution can be applied by completing the square.ExampleEvaluateZ dx.x 2 4x 13Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square.Sometimes we can convert an integral to a form where trigonometricsubstitution can be applied by completing the square.ExampleEvaluateZI dx.x 2 4x 13x 2 4x 13 x 2 2(2)x 22 22 13 (x 2)2 9.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square.Sometimes we can convert an integral to a form where trigonometricsubstitution can be applied by completing the square.ExampleEvaluateZII dx.x 2 4x 13x 2 4x 13 x 2 2(2)x 22 22 13 (x 2)2 9.RRR dx dx 2 du2 , where u x 2.2x 4x 13(x 2) 9Annette Pilkington(u) 9Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square.Sometimes we can convert an integral to a form where trigonometricsubstitution can be applied by completing the square.ExampleEvaluateZIII dx.x 2 4x 13x 2 4x 13 x 2 2(2)x 22 22 13 (x 2)2 9.RRR dx dx 2 du2 , where u x 2.2x 4x 13(x 2) 9(u) 9Now we apply the substitution u 3 tan θ, π2 θ π2 , du 3 sec2 θdθ,u 2 9 9 tan2 θ 9 3 sec2 θ 3 sec θ.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square.Sometimes we can convert an integral to a form where trigonometricsubstitution can be applied by completing the square.ExampleEvaluateZIIII dx.x 2 4x 13x 2 4x 13 x 2 2(2)x 22 22 13 (x 2)2 9.RRR dx dx 2 du2 , where u x 2.2x 4x 13(x 2) 9(u) 9Now we apply the substitution u 3 tan θ, π2 θ π2 , du 3 sec2 θdθ,u 2 9 9 tan2 θ 9 3 sec2 θ 3 sec θ.RRR2θdθ du 3 sec sec θdθ ln sec θ tan θ C 3 sec θ2(u) 9ln sec(tan 1 u3 ) tan(tan 1 u3 ) C .Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingIR dxx 2 4x 13 px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square. ln sec(tan 1 u3 ) tan(tan 1 u3 ) C , where u x 2.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving dx px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square. ln sec(tan 1 u3 ) tan(tan 1 u3 ) C , where u x 2.IRITo get an expression for sec(tan 1 u3 ), we use an appropriate trianglex 2 4x 13From the triangle, we getu 2 9sec(tan 1 u3 ) 3u2 9uθ3Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving dx px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the square. ln sec(tan 1 u3 ) tan(tan 1 u3 ) C , where u x 2.IRITo get an expression for sec(tan 1 u3 ), we use an appropriate trianglex 2 4x 13From the triangle, we getu 2 9sec(tan 1 u3 ) 3u2 9uθ3 IR dxx 2 4x 13 ln u 2 93 u3 C ln Annette Pilkington(x 2)2 93 Trigonometric Substitution(x 2) 3 C.

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a sec θ, 0 θ π2 or π θ 3π, ( This2amounts to saying θ sec 1 xa ), dx a sec θ tan θdθ, x 2 a2 a2 sec2 θ a2 a tan θ a tan θ (since 0 θ π2 or π θ 3πby choice)2ExampleEvaluateZ1 dxx 2 x 2 25Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a sec θ, 0 θ π2 or π θ 3π, ( This2amounts to saying θ sec 1 xa ), dx a sec θ tan θdθ, x 2 a2 a2 sec2 θ a2 a tan θ a tan θ (since 0 θ π2 or π θ 3πby choice)2ExampleEvaluateZI1 dxx 2 x 2 25Let x 5 sec θ 0 θ π2 or π θ 3π, then dx 5 sec θ tan θdθ,2 x 2 25 25 sec2 θ 25 5 tan2 θ 5 tan θ 5 tan θ.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involving px 2 a2 Integrals involvingx 2 a2x 2 a2We make the substitution x a sec θ, 0 θ π2 or π θ 3π, ( This2amounts to saying θ sec 1 xa ), dx a sec θ tan θdθ, x 2 a2 a2 sec2 θ a2 a tan θ a tan θ (since 0 θ π2 or π θ 3πby choice)2ExampleEvaluateZII1 dxx 2 x 2 25Let x 5 sec θ 0 θ π2 or π θ 3π, then dx 5 sec θ tan θdθ,2 x 2 25 25 sec2 θ 25 5 tan2 θ 5 tan θ 5 tan θ.RRR 1θ tan θ 1dx 25 5secsec2 θ(5dθ 25 secdθθtan θ)22xx 25 125Zcos θ dθ 11xsin θ C sin(sec 1 ) C25255Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingZ px 2 a2 Integrals involvingx 2 a211x dx sin(sec 1 ) C255x 2 x 2 25Annette PilkingtonTrigonometric Substitutionx 2 a2

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingZI px 2 a2 Integrals involvingx 2 a2x 2 a211x dx sin(sec 1 ) C255x 2 x 2 25To get an expression for sin(sec 1 x5 ), we use an appropriate triangleFrom the triangle, we getx 2 25sin(sec 1 x5 ) xxx2 -25θ5Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvingIntegrals involvingZIpa2 x 2 Integrals involving x 2 a2 Integrals involvingx 2 a2x 2 a211x dx sin(sec 1 ) C255x 2 x 2 25To get an expression for sin(sec 1 x5 ), we use an appropriate triangleFrom the triangle, we getx 2 25sin(sec 1 x5 ) xxx2 -25θ5 IHenceRx2 1x 2 25dx 125·x 2 25xAnnette Pilkington C x 2 2525xTrigonometric Substitution C.

Trigonometric Substitution Integrals involvingIntegrals involvingZIpa2 x 2 Integrals involving x 2 a2 Integrals involvingx 2 a2x 2 a211x dx sin(sec 1 ) C255x 2 x 2 25To get an expression for sin(sec 1 x5 ), we use an appropriate triangleFrom the triangle, we getx 2 25sin(sec 1 x5 ) xxx2 -25θ5 R 1·x 2 25x INote You can also use this substitution to getZ11x dx sec 1 C .aax x 2 a2x 2 25Annette Pilkington C x 2 2525xHencex2dx 125ITrigonometric Substitution C.

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleEvaluateR64 px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIEvaluateR64 px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Completing the square x 2 6x 8 x 2 2(3)x 9 9 8 (x 3)2 1Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIIEvaluateR64 px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Completing the square x 2 6x 8 x 2 2(3)x 9 9 8 (x 3)2 1RRR dxdx du , where u x 3.dx dxx 2 6x 8(x 3)2 1Annette Pilkingtonu 2 1Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involving Integrals involvingExampleIIIEvaluateR64 px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Completing the square x 2 6x 8 x 2 2(3)x 9 9 8 (x 3)2 1RRR dxdx du , where u x 3.dx dxx 2 6x 8(x 3)2 1Let u sec θ 0 θ π2u 2 1or π θ Annette Pilkington3π,2du sec θ tan θdθ.Trigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIIIIEvaluateR64 px 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Completing the square x 2 6x 8 x 2 2(3)x 9 9 8 (x 3)2 1RRR dxdx du , where u x 3.dx dxx 2 6x 8Let u sec θ 0 RR θ tan θdθ sec2sec θ 1(x 3)2 1θ π2 orsec θ tan θdθsec2 θ 1u 2 1π θ 3π, du sec θ tan θdθ.2R sec θdθAnnette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIIIIEvaluate4 x 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Completing the square x 2 6x 8 x 2 2(3)x 9 9 8 (x 3)2 1RRR dxdx du , where u x 3.dx dxx 2 6x 8Let u sec θ 0 RR θ tan θdθ sec2sec θ 1IR6 p(x 3)2 1θ π2 orsec θ tan θdθsec2 θ 1u 2 1π θ 3π, du sec θ tan θdθ.2R sec θdθ ln sec θ tan θ C ln sec(sec 1 u) tan(sec 1 u) Annette PilkingtonTrigonometric Substitution

Trigonometric Substitution Integrals involvinga2 x 2 Integrals involvingIntegrals involvingExampleIIIIEvaluateR64Ix 2 a2 Integrals involvingx 2 a2x 2 a2 , Completing the squaredxx 2 6x 8dx.Completing the square x 2 6x 8 x 2 2(3)x 9 9 8 (x 3)2 1RRR dxdx du , where u x 3.dx dx(x 3)2 1x 2 6x 8Let u sec θ 0 RR θ tan θdθ sec2sec θ 1I pθ π2 orsec θ tan θdθsec2 θ 1u 2 1π θ 3π, du sec θ tan θdθ.2R sec θdθ ln sec θ tan θ C ln sec(sec 1 u) tan(sec 1 u) To get an expression for tan(sec 1 u), we use an appropriate triangleuFrom the triangle, we get 1u 2 1 Hencetan(secu) R dxdxx 2 6x 8 ln u u 2 p1 C ln (x 3) (x 3)2 1 C .R6 dxdx4x 2 6x 8 ln 3 8 ln 1 ln 3 8 u2 -1θ1Annette PilkingtonTrigonometric Substitution

Trigonometric SubstitutionIntegrals involving q a2 x2 Integrals involving p x2 a2 Integrals involving q x2 a2 Integrals involving p a2 x