Show steps, thanks.
asked Apr 17, 2016 in Calculus Answers by anonymous

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Integral = I

I ( sec(2x))^3*tan(2x)dx =        , 2x = u,  2dx = du, dx = du/2

1/2* I (secu)^3*tanudu =  ,        secu = 1/cosu and tanu = sinu/cosu

1/2* I 1/(cosu)^3*sinu/cosudu =

1/2 I sinu/(cosu)^4* du =                  cosu=v, - sinudu =dv, and sinudu = - dv

1/2 I - dv/v^4 =

-1/2I v^(-4) dv =

-1/2 * v^(-4+1)/(-4+1) =

1/6*v^-3 = 1/(6v^3) = 1/(6(cosu)^3) =

1/6* 1/((cos(2x))^3 + C

 OR

2x = u

dx = du/2

= 1/2*I sec^2u*secu*tanu du

secu =v and secu*tanu du=dv, sec^2u=v^2

!/2I v^2 dv =

1/6*v^3 = 1/6* sec^3u = 1/6*1/cos^3(2x) + C
answered Apr 17, 2016 by jkhenry Level 8 User (36,160 points)
edited Apr 17, 2016 by jkhenry
Let p=cos(2x) then dp=-2sin(2x)dx, so sin(2x)dx=-dp/2

sec^3(2x)tan(2x)dx=sin(2x)/cos^4(2x)dx=-dp/(2p^4).

-(1/2)S(p^-4dp), where S denotes integral, =(1/2)p^-3/3=(1/6)sec^3(2x).

CHECK: Differentiate sec^3(2x)/6: (6/6)sec^2(2x)sec(2x)tan(2x)dx=sec^3(2x)tan(2x)dx. OK!
answered Apr 17, 2016 by Rod Top Rated User (415,140 points)
Yep, I forgot the constant of integration. Thanks, jkhenry, for reminding me!

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