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Engineering Calculus II Week 5 Notes

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by: Kevin Notetaker

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Engineering Calculus II Week 5 Notes MATH 2152

Marketplace > East Carolina University > Mathematics (M) > MATH 2152 > Engineering Calculus II Week 5 Notes
Kevin Notetaker
ECU

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Notes from Dr. Guglielmo Fucci's Engineering Calculus II Class
COURSE
Engineering Calculus II
PROF.
Dr. Guglielmo Fucci
TYPE
Class Notes
PAGES
6
WORDS
KARMA
25 ?

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1 review
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Popular in Mathematics (M)

This 6 page Class Notes was uploaded by Kevin Notetaker on Wednesday March 9, 2016. The Class Notes belongs to MATH 2152 at East Carolina University taught by Dr. Guglielmo Fucci in Spring 2016. Since its upload, it has received 65 views. For similar materials see Engineering Calculus II in Mathematics (M) at East Carolina University.

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Date Created: 03/09/16
3/2/2016 Trigonometric Substitutions (Sect. 7.2)  Things to know:  sin(x)+cos2(x=1∧1+tan 2(x=sec 2(x)  This allows you to compute the integrals containing the following expressions.  b −x usesin (u) x +b usetan (u)  √ √ 2 2 1  √ x −b usesec(u)∨ cos(u) ≠  b is any real number   0 b −x dx  ∫ √  Substitute bsin(u) for the value of x  x = bsin(u)  dx = bcos(u)du  ∫ √b −b sin (u)  bcos(u)du  Factor out the  b2 1−sin (u) ¿ b ¿  √ ¿ ∫ ¿ 2 2 2 2  Since  sin x +cos (x)=1;    cos (x=1−sin (x) 2 2 2  ∫ b√cos (u)bcos(u)du=b √cos (u)cos(u)du  Simplify the equation b2∫ cos ( )u   Compute Integral 2 2 b [ cos(u)sinu + ] b u+ b cos(usin(u)+C  2 2 ;  2 2  Substitute x back x=¿arcsin x =u  sin(u) =  a (a 3/2/2016 cos arcsin = 1−sin arcsinx  ( (b ) √ ( (b ); ArcSin is the derivative of Sin. So the  cancel out x2 bx x2  1− b2 ; 2 1−b2 √ √  You can then factor out 1/ bx 1√b −x = x√b −x2  (2 b 2  So 2  ∫ √b −x =b arcsinx+ x√b −x2 2 (b 2 3/2/2016 2 2  ∫ √ x +b dx  Substitute btan(x) for values of x  x = btan(u) b  dx =  2 du cos (u) b 1 ∫ √b tan2u)+b2 2 du=b ∫ √ tanu +1 2 du  cos ( ) cos( ) 1+tan2u =sec2(u∨ 1  Since cos (u) , plug it back into the integral. 2 1 1 2 1  b ∫ cos ( )os u( )du=b ∫ cos (u)du √ 1 1  Since  ∫ cos(x)dx=ln ∨ +tan(x)∨¿   then factor out a 1/cos(x) to  co( ) simplify the problem. tanu ) 2 2 ∗1 1 1  b ( +2 ∫ du)    cos(u) cosu)  Now simplify the problem and compute the integral 2 b ∗tanu)  2 b 2 1 cosu ) + 2 l∨?cos(u)tan (u)∨+C  Substitute x back tanu = ;u=arctan( )  b b 2 1 2 x x  x = √+tan (arcta(b );whichsimplifie√¿ 1b2 cos(arct() ) b  So 2 b ∗x 2 2 2 x2 b2 x x2  ∫ √b +x = b 1+ 2+2 ln|b+ 1+ 2|C √ b √ b  Which Simplified  3/2/2016 2 2 x+ √ +x ∨2C  =  x 2 2 b 1 2 √b +x + 2 ln∨ b 3/2/2016 x −b dx  ∫ √ b  Substitute   for values of x cos( ) b  x =  cos(u) b  dx =  cos(u)tan(u)du 2 tan(u)  ∫ 2 −b 2 b tanu)du=b2∫ 12 −1 du √cos (u) cos(u) √cos u cos(u) 2 −1=tan (u)  Since  cos (u)  then plug it back into the integral. b2∫√ tan (u)an(u)du=b ∫ 1 tan (u)du  cos(u) cos(u) 1 −1=tan (u)  Since  cos (u) , Substitute it back in for the integral b2 1 1 −1 du=b 2 1 − 1 du  ∫ cosu)(cos(u) ) ∫ cos (u) cos(u)  Now we can separate each integral    2tanu)  b2∫ 12 1 du−b 2∫ 1 du= b − ln?∨ 1 +tan(u)∨+C cos (u)os(u) co( ) 2 cos( ) 2 cos( )  Substitute x Back 1 = ;u=arccos( )  cos(u) b b x 1 x2 tan(arcco(b ) x −1;whichsimplifiesb2−1  cos(arcco() ) √ √ b  So 2 b  2 2 2∗x 2 2 b2 x 1 2 2 ∫ √x −b dx= 2 √x −b − 2 l|b b √x −b |C b  Which Simplified 3/2/2016 2 2 x+√x −b 2C  = x 2 2 b 1 2 √x −b − 2 ln∨ b

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