T1.14 Integral: trigonometric u-substitution

[karush]

$\tiny{2214.t1.14}$
$\text{Evaluate the Integral:}$
\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

 

[MarkFL]

Re: t1.14 Integral: trigonometic u-substitution

My suggestion, again, is not to use so many abbreviations in thread titles. (Yes)