MHB T1.14 Integral: trigonometric u-substitution

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The integral I_{14} is evaluated using the substitution u = 4 + tan^3(x), leading to dx expressed in terms of du. The integral simplifies to 4 times the integral of 1/u^2, resulting in -4/u after integration. Substituting back gives the final result as I_{14} = -4/(4 + tan^3(x)) + C. The discussion highlights the effectiveness of u-substitution in solving trigonometric integrals. Suggestions include avoiding excessive abbreviations in thread titles for clarity.
karush
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$\tiny{2214.t1.14}$
$\text{Evaluate the Integral:}$
\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
 
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Re: t1.14 Integral: trigonometic u-substitution

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

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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