$\displaystyle \[\int tan^3(x) sec^4(x) dx\] $ I first tried substituting $\displaystyle \[u = tan(x)\]$, then I tried substituting $\displaystyle \[u = sec(x)\]$, but I couldn't seem to solve it. Any hints?
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Originally Posted by BrownianMan $\displaystyle \[\int tan^3(x) sec^4(x) dx\] $ I first tried substituting $\displaystyle \[u = tan(x)\]$, then I tried substituting $\displaystyle \[u = sec(x)\]$, but I couldn't seem to solve it. Any hints? $\displaystyle \int tan^3(x)sec^4(x)$ cab be written as $\displaystyle \int tan^3(x)sec^2(x)sec^2(x)$ $\displaystyle \int tan^3(x)[1 + tan^2(x)]sec^2(x)$ Now simplify and solve.
$\displaystyle tan^2(x)+1=sec^2(x)$
Thank you!
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