- #1

karush

Gold Member

MHB

- 3,269

- 5

$\text{Evaluate the Integral:}$

\begin{align*}\displaystyle

I_{14}&=\int \frac{12\tan^2x \sec^2 x}{(4+\tan^3x)^2} \, dx \\

\textit{Use U substitution}&\\

u&=4+\tan^3x\\

\, \therefore dx& =\dfrac{1}{3\sec^2\left(x\right)\tan^2\left(x\right)}\,du\\

&=4 \int\frac{1}{u^2}\,du\\

&=4\left[-\dfrac{1}{u} \right]\\

\textit{Back substitute $u=4+\tan^3x$}\\

I_{14}&=-\frac{4}{4+\tan^3x}+C

\end{align*}

ok just seeing if this is correct

and suggestions