How do you integrate? $\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$

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How do you integrate? $\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$

$\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$

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Answer 1 · 2018-04-14T19:29:44

$\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \sqrt{x^{2} - 1}} = \int_{\frac{π}{4}}^{\frac{π}{3}} {cos}^{4} θ d θ$ $int_sqrt2^2dx/(x^5sqrt(x^2-1))=int_(pi/4)^(pi/3)cos^4thetad theta$ , which I leave to you to solve.

Explanation:

Try letting $x = sec θ$ $x=sectheta$ . Note this implies that $d x = sec θ tan θ d θ$ $dx=secthetatanthetad theta$ .

Also, the bounds will change. Observe that $x = \sqrt{2} \Rightarrow θ = \frac{π}{4}$ $x=sqrt2=>theta=pi/4$ and $x = 2 \Rightarrow x = \frac{π}{3}$ $x=2=>x=pi/3$ .

Then:

$\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \sqrt{x^{2} - 1}} = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{sec θ tan θ d θ}{{sec}^{5} θ \sqrt{{sec}^{2} θ - 1}}$ $int_sqrt2^2dx/(x^5sqrt(x^2-1))=int_(pi/4)^(pi/3)(secthetatanthetad theta)/(sec^5thetasqrt(sec^2theta-1))$

Recall the trigonometric identity ${sec}^{2} θ - 1 = {tan}^{2} θ$ $sec^2theta-1=tan^2theta$ .

$= \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{sec θ tan θ d θ}{{sec}^{5} θ tan θ}$ $=int_(pi/4)^(pi/3)(secthetatanthetad theta)/(sec^5thetatantheta)$

$= \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d θ}{{sec}^{4} θ}$ $=int_(pi/4)^(pi/3)(d theta)/sec^4theta$

$= \int_{\frac{π}{4}}^{\frac{π}{3}} {cos}^{4} θ d θ$ $=int_(pi/4)^(pi/3)cos^4thetad theta$

From here, I would use the identity ${cos}^{2} α = \frac{1}{2} (1 + cos 2 α)$ $cos^2alpha=1/2(1+cos2alpha)$ , or, in this case, ${cos}^{4} α = \frac{1}{4} {(1 + cos 2 α)}^{2}$ $cos^4alpha=1/4(1+cos2alpha)^2$ to further simplify the integral. It will definitely be a little messy.

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How do you integrate? $\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$

$\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$

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Explanation:

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How do you integrate? int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))∫2√2dxx5⋅√x2−1int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))

int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))∫2√2dxx5⋅√x2−1int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))

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Explanation:

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How do you integrate? $\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$

$\int_{\sqrt{2}}^{2} \frac{d x}{x^{5} \cdot \sqrt{x^{2} - 1}}$ $int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))$