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## Calculus II Week 2 Notes

by: Blanche Langenbach

8

0

7

# Calculus II Week 2 Notes MAC 2312

Marketplace > University of South Florida > Mathematics > MAC 2312 > Calculus II Week 2 Notes
Blanche Langenbach
USF
GPA 4.0

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Week 2 Day 1 and 2 Notes August 29 and 31 Integration by parts with trigonometry
COURSE
Calculus II
PROF.
David Lee Milligan
TYPE
Class Notes
PAGES
7
WORDS
CONCEPTS
Integration, by, parts, Trigonometry, Calculus, calculus2
KARMA
25 ?

## Popular in Mathematics

This 7 page Class Notes was uploaded by Blanche Langenbach on Friday September 2, 2016. The Class Notes belongs to MAC 2312 at University of South Florida taught by David Lee Milligan in Fall 2016. Since its upload, it has received 8 views. For similar materials see Calculus II in Mathematics at University of South Florida.

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Date Created: 09/02/16
Week 2 Day 1 Notes 8/29/16 – 8/31/16 6.2 Trigonometric Integrals and Substitutions Calculus II *Important to know trigonometric identities for this section* Using the Pythagorean identity ∫ sin ( )os x( ) Ex 1:  sin2(x)=1−cos 2(x ) sin (¿¿2x) cos (x)sin x )dx u=cos (x) ∫ ¿ du=−sin (x)dx 2 cos (x) 1−¿ −du=sin (x)dx ¿ ¿ ∫ ¿ 2 u 1−¿ ¿ ¿ ∫ ¿ 2 4 6 −∫ (u −2u +u )du 3 5 7 − u −2 u +u +C (3 5 7) −1 cos3(x− cos (x)− cos 7(x)+C 3 5 7 Using the Half­Angle Identities 4 Ex 2:  ∫ sin( )dx (x) sin ¿ ¿ sin ( )¿ 2 1 sin x = 21−cos 2 ( ) ) Half­Angle Identity 1−cos 2x ) 2 1 2 cos ( x ) (1+cos (4 x)) ¿ 2 ¿ ¿ ∫ ¿ 1 2 4 ∫ (−2 cos 2(x )cos 2(x d)) 1 3 1 4 ∫ (2−2cos 2(x +)c2s? (4x))dx Finish the Integration for practice Using Pythagorean Identity 5 7 Ex3:  ∫ tan ( )ec x( ) 7 6 5 4 Separate sec (x) into sec(x) and sec (x) and tan (x) into tan (x) and tan(x) tan 2 Separate sec(x)tan(x) fr❑m rema2(x)=secpr(x−1ons and convert power of tangent to expression  involving secant using: 4 6 ∫ t an( )sec x( )c x( )n x( ) 2 sec u=sec (x) (¿¿2 x −1) sec 6(x)sec(x tan(x dx ¿ ∫ du=sec (x)tan(x)dx 2 2 6 ∫ (u −1) u du 10 8 6 ∫ (u −2u +u )du 11 9 7 u u u 11 −2 9 + 7 +C sec (x) 2 9 1 7 − sec x + sec x +C 11 9 7 Formulas to Know: ∫ tan( )=ln∨sec?(x)∨+C ∫ sec( )=ln|secx+tanx|+C Integrate by Parts with Trig 3 u=sec x)dv=sec (x)dx Ex 4:    ∫ sec( )dx   du=sec x tan(x)dxv=tan ?(x) sec( )tan( )− sec ( )an x( ) =  ∫ sec(x)tan(x− ∫ec (x)(sec(x)−1 )x =  = sec ( )an( )−∫sec x( )+ ∫ sec( )dx sec x tan(x)+ln?∨sec x +tan?(x)∨+C 1 = sec(x)tan(x− ¿ 2 Inverse Trigonometric Substitutions √a −x 2 x=asin? (x) −π π [ ] , 2 2 2 2 √a +x x=atan ?(x) ⌊−π π, ⌋ 2 2 √x −a 2 x=asec ?(x) 0≤ x< π∨π≤x< 3π 2 2 Finding area enclosed by ellipse:  2 2 x y a2+ b2=1 [0, a] 1  Quadrant b/c of symmetric on both axes x=a  sinθ b 2 2 y= a √a −x b 1 A=∫ √a −x dx 4 a Change limits:  x=0, sin (θ )= 0,  θ =0 θ¿=1,θ= π x=1, sin ( 2 π 2 4b acosθ(acosθ dθ a 0 Area of ellipses with semi­axes a and b =  2 If a=b=r, then area of circle is proven as  dx a>0 Ex 5:  ∫ √x −a 2 Let x=a  secθ       dx= a ecθtanθdθ 2 2 2 2 √x −a = a√(sec θ−1) 2 2      = a tan θ 2 2 √ x −a      = |tanθ=atanθ tanθ= a asecθtanθ Ex 6:  ∫ dθ √x −a 2 atanθ ∫ secθdθ x secθ= x a ln¿secθ+tanθ∨+C C =C−ln ? (a) 1 x √ x −a 2 θ ln¿ + | + C a a a 2 2 ln x+ √ −a +C| 1 *If you are unsure of what to define as u and dv in integrating by parts, use: LIATE to choose ‘u’ to be the function that comes first in this list L­logarithmic function I­inverse trig function A­algebraic function T­trigonometric function E­exponential function

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