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Problem: $\int \sqrt{150+25x^2}\ dx $

The only way I've solved this problem before was with hyperbolic integral rules and substitution, but I'm supposed to solve it with integral trig substitution. Not quite sure where to start..

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Hint: You can pull out the factor $25$ from the square root, then define $u=ax$ for a suitable $a$ to make the square root $\sqrt {1+u^2}$ Then do you know any trig functions that when squared and added to one make a perfect square? Set $u$ equal to that and figure out what $du$ is.

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$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\expo}{{\rm e}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$

$$ \overbrace{\int\sqrt{\vphantom{\Large A}150 + 25x^{2}\,}\,\dd x} ^{x\ =\ \sqrt{\vphantom{\large A}6\,}\,\tan\pars{\theta}}\ =\ 30\int\sec^{3}\pars{\theta}\,\dd\theta $$
\begin{align} \int\sec^{3}\pars{\theta}\,\dd\theta &= \int\overbrace{\sec^{2}\pars{\theta}} ^{\tan^{2}\pars{\theta}\ +\ 1}\sec\pars{\theta}\,\dd\theta = \int\tan\pars{\theta}\ \overbrace{\tan\pars{\theta}\sec\pars{\theta}\,\dd\theta} ^{\dd\bracks{\sec\pars{\theta}}} + \int\sec\pars{\theta}\,\dd\theta \\[3mm]&= \tan\pars{\theta}\sec{\theta} - \int\sec^{3}\pars{\theta}\,\dd\theta + \int\sec\pars{\theta}\,\dd\theta \end{align}
\begin{align} \int\sec^{3}\pars{\theta}\,\dd\theta &= {1 \over 2}\bracks{% \tan\pars{\theta}\sec{\theta} + \int\sec\pars{\theta}\,\dd\theta} = {\tan\pars{\theta}\sec{\theta} + \ln\pars{\sec{\theta} +\tan\pars{\theta}} \over 2} \\[3mm]&= {\tan\pars{\theta}\sqrt{\tan^{2}{\theta} + 1\,} + \ln\pars{\sqrt{\tan^{2}{\theta} + 1\,} +\tan\pars{\theta}} \over 2} \\[3mm]&= {\pars{x/\sqrt{6\,}}\sqrt{\pars{x/\sqrt{6\,}}^{2} + 1\,} + \ln\pars{\sqrt{\pars{x/\sqrt{6\,}}^{2} + 1\,} + x/\sqrt{6\,}} \over 2} \\[3mm]&= {x\,\sqrt{x^{2} + 6\,} + 6\ln\pars{\bracks{\sqrt{x^{2} + 6\,} + x}/\sqrt{6\,}} \over 12} \\[3mm]&= {x\,\sqrt{x^{2} + 6\,} + 6\ln\pars{\sqrt{x^{2} + 6\,} + x} - 3\ln\pars{6} \over 12} \end{align}

$$\color{#ff0000}{\large% \int\sqrt{\vphantom{\Large A}150 + 25x^{2}\,}\,\dd x \color{#000000}{\ =\ } {5 \over 2}\bracks{% x\,\sqrt{\vphantom{\Large A}x^{2} + 6\,} + 6\ln\pars{\sqrt{\vphantom{\Large A}x^{2} + 6\,} + x}} + \color{#0000ff}{\mbox{constant}}} $$

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