Integrals Related to Inverse Trig, Inverse Hyperbolic Functions

Integrals Related to Inverse Trig, Inverse Hyperbolic Functions

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 dx u u 2 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 When given given integral integral problems, problems, look look for for these these patterns 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 x dx 1 x 2 dx 1 x 2 1 x 2 dx 1 x

2 IfIf they they are are not, not, how how are are they they integrated? integrated? 5 Simplify. 1. 2. x dx 4 x2 dx 1 2 4 x 64

dx The hardest part is getting the integral into the proper form. dx 4 x 2 a 2 du du 2 a u 2 sin 1 ua C u x du dx

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 du 2

a u 2 sin 1 ua C u x du dx 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 2 a u a 1 1 du 2 a2 u2 1 1 u sin a C 2 2 sin 1 ua C 2 u x du 2 xdx 1 1 2 sin x C 2 The hardest part is getting the integral into the proper form. dx

3 4x 2 du 2 a u a 3 1 du 2 a2 u2 1 1 u sin a C 2 2 sin 1 ua C u 2 x du 2dx 1 1 2x sin C 2 3

The hardest part is getting the integral into the proper form. 1 dx 2 1 x 0 du 1 1 u a 2 u 2 a tan a C a 1 u x du dx 1 tan x 1 0 tan 1 1 tan 1 0 0

4 4 Simplify. 1. 2. x dx 4 x2 dx 1 2 4 x 64 dx Simplify. 1. 2. dx

x 4 x 2 dx arcsin 2 C a 2 u x du dx x 1 dx 4 x 64 x 2 1 2 2x 8 u 2 x du 2dx 2x 2 dx a 8 1

2x 2 8 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 2 a 4 4. 1

9 2x 3 2 dx t 2 2 1 dt 2 u t 2 du 2tdt 3 a 3 2x 3 4 2 t

2 2 dt t2 1 arcsin C 2 4 1 2 2t 2 dx u 2 x 3 du 2dx 1 1 2 2dx 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 2 x 10 x 16 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 dx x 1 2 9 du u 2 a 2 COMPLETE THE SQUARE!!!! u x 1 du dx a 3 1

x 1 tan 1 C 3 3 x 2 2 x __ 10 __ x 2 2 x 1 10 1 x 1 2 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 1 x 3 arctan C 6 6 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

du dx a 1 1 arctan x 2 C 2 3. dx 3x x2 3. dx 3x x 2 dx 2 x 3x

dx x 2 3x dx x 2 3 x 94 94 dx 9 4 x 2 3 x 94 dx 3 2 2 x x arcsin 3 2 3 2 3 2 2

C 2x 3 arcsin C 3 a 32 u x du dx 3 2 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 2x 3 1 1 dx

u 4 x x 2 du 4 2 x dx 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 x 1 dx u x 2 1. dx 2 2udu x 1 u 2 2 1 u x 2 u 2 2 du u 2 2 x u 3 2udu dx u2 3 3 2 2 du u 3 u2 3 1 2 2 du 6 2 du u 3 u 3 1

2du 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 dx dx 2 2 x 2x 1 5 1 x 2x 5 1 1 1 0.5 3 2 1 2

fnInt 1/ x 2 x 5 , x,1,3 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 2 dx 1 x

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