Evaluate $\int\frac{\sqrt {25 - x^2}}{ x^4}$

Question

I'm pretty sure the method used is trig substitution. But I'm having trouble setting up and solving the problem.

Answer 1

Subsititute $\theta=\sin^{-1} \dfrac{x}5\implies \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{25-x^2}}$, so we get,

\begin{align} \\\\\\&\int\frac{\sqrt{25-x^2}}{x^4}dx \\=&\int\frac{25-x^2}{x^4}\dfrac{1}{\sqrt{25-x^2}}d\theta \\=&\int \dfrac{25-25\sin^2 \theta}{625\sin^4\theta}d\theta \\=&\dfrac{1}{25}\int\dfrac{\cos^2 \theta}{\sin^4\theta}d\theta \\=&\dfrac{1}{25}\int\cot^2\theta \csc^2\theta d\theta \end{align}

Now subsitute $u=\cot \theta$ to finish.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{x \equiv {1 \over u}}$: \begin{align} \color{#00f}{\large\int{\root{25 - x^{2}} \over x^{4}}\,\dd x} &=\int{\root{25 - 1/u^{2}} \over 1/u^{4}}\, \pars{-\,{\dd u \over u^{2}}} =-\int u\root{25u^{2} - 1}\,\dd u \\[3mm]&=-\,\half\int\root{25u^{2} - 1}\,\dd\pars{u^{2}}=-\,{\pars{25u^{2} - 1}^{3/2} \over 75} =-\,{\bracks{25\pars{1/x}^{2} - 1}^{3/2} \over 75} \\[3mm]&=\color{#00f}{\large -\,{\root{25 - x^{2}} \over 75x^{3}}} + \pars{~\mbox{a constant}~} \end{align}