integrating with substitution

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integrate sec 1/2x(sec 1/2x+tan1/2x)dx.

What I did.......... I multiplied out and got (sec 1/2x)^2+sec 1/2x tan1/2x dx
then used (sec 1/2x)^2 for u. left over is sec1/2xtan1/2x. du=sec1/2xtan1/2x

integrate u du and get u^2/2 substitute back in and get (sec 1/2x)^4/2

When I attempt to check my work with the derivative, I appear to be wrong in my answer. Can anyone tell me where I went wrong?

Thanks
 
Are the "1/2x" parts meant to be "1/(2x)" or "(1/2)x"?

Thank you.

Eliz.
 
Is this your integral?.

\(\displaystyle \L\\\int{sec(\frac{x}{2})\left[sec(\frac{x}{2})+tan(\frac{x}{2})\right]\)
 
the bottom display of the integral is correct. it is sec(1/2(x)) and tan of the same thing Thanks
 
You are correct. You have:

\(\displaystyle \L\\\int{sec^{2}(\frac{x}{2})}dx+\int{sec(\frac{x}{2})tan(\frac{x}{2})}dx\)

You should be able to find these in a table of integrals.

\(\displaystyle \L\\\int{sec^{2}(\frac{x}{2})}dx\)

Let \(\displaystyle u=\frac{x}{2},\;\ 2du=dx\)

\(\displaystyle \L\\\int{sec^{2}(u)}du=2tan(\frac{x}{2})\)


\(\displaystyle \L\\\int{sec(\frac{x}{2})tan(\frac{x}{2})}dx=2sec(\frac{x}{2})\)
 
Thanks

WOW was I making that harder then need be!! I guess I am trying to complicate things. Thanks for the help! :D
 
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