mronjo7n

2021-11-17

Use trigonometric substitution to integrate (Don’t forget to write your answer in terms of the original variable xand do not leave trigonometric functions evaluated at inverse trigonometric functions)
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx$

Ralph Lester

Step 1
Given:
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx$
To find:
Use trigonometric substitution to integrate above function.
Step 2
Apply trigonometric substitution: $x=\mathrm{sec}\left(u\right)$
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\int \frac{\sqrt{{\mathrm{sec}}^{2}\left(u\right)-1}}{{\mathrm{sec}}^{4}\left(u\right)}du$
$=\int \frac{{\mathrm{tan}}^{2}\left(u\right)}{{\mathrm{sec}}^{3}\left(u\right)}du$
Rewrite using trigonometric identities,
$\mathrm{tan}\left(u\right)=\frac{\mathrm{sin}\left(u\right)}{\mathrm{cos}\left(u\right)}$ and $\mathrm{sec}\left(u\right)=\frac{1}{\mathrm{cos}\left(u\right)}$
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\int {\mathrm{sin}}^{2}\left(u\right)\mathrm{cos}\left(u\right)du$
Substitute $v=\mathrm{sin}\left(u\right)$, and $dv=\mathrm{cos}\left(u\right)du$
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\int {v}^{2}dv$
using the formula, $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\frac{{v}^{2+1}}{2+1}$
On back substitution,
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\frac{{\mathrm{sin}}^{3}\left(u\right)}{3}$
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\frac{{\mathrm{sin}}^{3}\left({\mathrm{sec}}^{-1}\left(x\right)\right)}{3}$
On simplifying above function,
$\int \frac{\sqrt{{x}^{2}-1}}{{x}^{4}}dx=\frac{{\left({x}^{2}-1\right)}^{\frac{3}{2}}}{3{x}^{3}}$

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