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Mathematics 41 Online
OpenStudy (anonymous):

complete the following integral or partial substitution with x tanx dx

OpenStudy (dumbcow):

use integration by parts u = x dv = tanx du =dx v = -ln|cosx|

OpenStudy (dumbcow):

wait nevermind that wont work i dont know if this has a simple anti-derivative

OpenStudy (anonymous):

Do integration by parts, Take the derivative of x and integral of tanx =\[x \log|secx| - \int\limits_{}{}tanx\] =log|secs|(x-1)

OpenStudy (anonymous):

That should be secx*

OpenStudy (anonymous):

Sorry did a mistake, its not that simple, will think on it!

OpenStudy (anonymous):

= uv - integral v du so now all thats left is finding integral of ln(secx)

OpenStudy (anonymous):

let u= sec(x) , du = sec(x)tan(x) dx --> dx = du/ sec(x)tan(x)

OpenStudy (anonymous):

so integral ln(u) du / sec(x) tan(x) = \[\frac{ \ln(u) du }{u \sqrt{u^2-1}} \]

OpenStudy (anonymous):

yes , hmm, I dont think you can go anywhere there

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