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complete the following integral or partial substitution with x tanx dx
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use integration by parts u = x dv = tanx du =dx v = -ln|cosx|
wait nevermind that wont work i dont know if this has a simple anti-derivative
Do integration by parts, Take the derivative of x and integral of tanx =\[x \log|secx| - \int\limits_{}{}tanx\] =log|secs|(x-1)
That should be secx*
Sorry did a mistake, its not that simple, will think on it!
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= uv - integral v du so now all thats left is finding integral of ln(secx)
let u= sec(x) , du = sec(x)tan(x) dx --> dx = du/ sec(x)tan(x)
so integral ln(u) du / sec(x) tan(x) = \[\frac{ \ln(u) du }{u \sqrt{u^2-1}} \]
yes , hmm, I dont think you can go anywhere there
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