Inverse Trigonometric Functions: Integration Lesson 5.8 Review • Recall derivatives of inverse trig functions d 1 du 1 sin u , u 1 dx 1 u 2 dx d 1 du 1 tan u dx 1 u 2 dx d 1 du 1 sec u , u 1 2 dx u u 1 dx 2 Integrals Using Same Relationships du u a 2 u 2 arcsin a C du 1 u a 2 u 2 a arctan a C du 1 u u u 2 a 2 a arcsec a C When given integral problems, look for these patterns 3 Identifying Patterns • For each of the integrals below, which inverse trig function is involved? 4dx 13 16 x 2 x dx 25 x 2 4 dx 9 x 2 4 Warning • Many integrals look like the inverse trig forms • Which of the following are of the inverse trig forms? x dx 1 x2 dx 1 x2 x dx 1 x2 dx 1 x 2 If they are not, how are they integrated? 5 Simplify. 1. 2. x dx 4 x2 dx 1 4 x 64 2 dx The hardest part is getting the integral into the proper form. dx 4 x 2 a2 du a u 2 2 sin 1 ua C ux du dx du a2 u2 sin C 1 u a x sin C 2 1 The hardest part is getting the integral into the proper form. dx 9 x 2 a 3 du a u 2 2 sin 1 ua C ux du dx du a2 u2 sin C 1 u a x sin C 3 1 The hardest part is getting the integral into the proper form. xdx 1 x 4 du a u a 1 1 du 2 a2 u2 1 1 u sin a C 2 2 2 sin 1 ua C ux du 2xdx 2 1 1 2 sin x C 2 The hardest part is getting the integral into the proper form. dx 3 4x 2 du a u a 3 1 du 2 a2 u2 1 1 u sin a C 2 2 2 sin 1 ua C u 2x du 2dx 1 1 2 x sin C 2 3 The hardest part is getting the integral into the proper form. 1 dx 0 1 x 2 du 1 1 u a 2 u 2 a tan a C a 1 ux du dx 1 tan x 1 0 tan 1 1 tan 1 0 4 0 4 Simplify. 1. 2. x dx 4 x2 dx 1 4 x 64 2 dx Simplify. 1. 2. x 4 x 2 dx arcsin 2 C a2 ux du dx dx x 1 4 x 64 2 dx 1 2x 8 2 2 u 2x du 2dx a8 x dx 1 2x 2x 8 2 2 2dx x 2x 1 1 arc sec C arc sec C 8 8 4 8 Simplify. 3. 4. t 16 t 4 dt 1 9 2x 3 2 dx Simplify. 3. t 16 t 4 dt t 4 t 2 a4 4. 1 9 2x 3 2 dx 2 2 1 dt 2 u t2 du 2tdt 3 a3 2x 3 4 2 t 2 2 dt t2 1 arcsin C 2 4 1 2 2t 2 dx u 2x 3 du 2dx 1 1 2dx 2 2 2 3 2x 3 1 2x 3 arctan C 6 3 Try These • Look for the pattern or how the expression can be manipulated into one of the patterns 8dx 1 16 x 2 x dx 1 25 x 2 dx 4 x 2 4 x 15 x 5 x 10 x 16 2 16 dx Completing the Square • Often a good strategy when quadratic functions are involved in the integration dx x 2 2 x 10 • Remember … we seek (x – b)2 + c – Which might give us an integral resulting in the arctan function Example dx x 2 2 x 10 COMPLETE THE SQUARE!!!! dx x 12 9 u x 1 du u2 a2 a 3 du dx 1 x 1 tan 1 C 3 3 x x 2x 1 10 1 2 2 x __ 10 __ 2 x 12 9 1. dx x 2 6 x 15 2. dx 2x 2 8 x 10 1. dx 1 dx x 2 6 x 15 x 2 6 x 9 15 9 x 3 2 6 dx u x 3 a 6 du dx x 3 arctan C 6 6 1 2. dx 1 1 1 1 2x 2 8 x 10 2 x 2 4 x 5 dx 2 x 2 4 x 4 5 4 dx 1 1 dx 2 2 x 2 1 u x 2 a 1 du dx 1 arctan x 2 C 2 3. dx 3x x 2 3. dx 3x x 2 dx x 3x 2 dx x 2 3x dx x 2 3 x 94 94 dx 9 4 x 2 3 x 94 dx x 3 2 2 3 2 2 x 32 arcsin 3 C 2 2x 3 arcsin C 3 a 32 u x 32 du dx Rewriting as Sum of Two Quotients • The integral may not appear to fit basic integration formulas – May be possible to split the integrand into two portions, each more easily handled 4x 3 1 x 2 dx 4x 1 x 2 3 1 x 2 4. 2x 3 4x x 2 dx 4. 2x 3 4x x 2 dx u 4x x 2 du 4 2x dx 2x 3 1 1 dx 4x x 2 2x 4 1 dx dx 4x x 2 4x x 2 2x 4 1 dx dx 2 2 4x x x 4x 1 12 u du dx x 2 4x 4 4 1 12 u du dx a 2 2 u x 2 4 x 2 du dx 1 x 2 2 2u arcsin C 2 1 x 2 2 2 2 4 x x arcsin C 2 Basic Integration Rules • Note table of basic rules • Most of these should be committed to memory • Note that to apply these, you must create the proper du to correspond to the u in the formula cos u du sin u C 1. x 2 dx x 1 u x 2 1. dx 2 2udu x 1 u 2 1 u x 2 u2 2 2 du u2 2 x u 3 2udu dx u2 3 3 2 du 2 u 3 u2 3 1 2 2 du 6 2 du u 3 u 3 1 2 du 6 2 du u 3 1 u 2u 6 arctan C 3 3 x 2 6 2 x 2 arctan C 3 3 2. Find the area of the region bounded by 1 y 2 , y 0, x 1, and x 3. x 2x 5 2. Find the area of the region bounded by 1 y 2 , y 0, x 1, and x 3. x 2x 5 3 3 1 1 1 x 2 2x 5 dx 1 x 2 2x 1 5 1dx 1 0.5 3 2 1 fnInt 1/ x 2x 5 , x,1,3 2 1 x 1 2 4 dx 3 1 x 1 arctan 2 1 2 1 1 arctan 1 arctan 0 2 2 1 0 2 4 8 .251? 3. dy xy dx 1 x 2