How do you integrate $\int \frac{1}{x^{4} \sqrt{16 + x^{2}}}$ $int 1/(x^4sqrt(16+x^2))$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{x^{4} \sqrt{16 + x^{2}}}$ $int 1/(x^4sqrt(16+x^2))$ by trigonometric substitution?

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Answer 1 · 2018-03-23T14:21:23

The answer is $= \frac{\sqrt{16 + x^{2}}}{256 x} - \frac{{(x^{2} + 16)}^{\frac{3}{2}}}{768 x^{3}} + C$ $=sqrt(16+x^2)/(256x)-(x^2+16)^(3/2)/(768x^3)+C$

Explanation:

Let $x = 4 tan θ$ $x=4tantheta$ , $\Rightarrow$ $=>$ , $d x = 4 {sec}^{2} θ d θ$ $dx=4sec^2thetad theta$

$\sqrt{16 + x^{2}} = \sqrt{16 + 16 {tan}^{2} θ} = 4 sec θ$ $sqrt(16+x^2)=sqrt(16+16tan^2theta)=4sectheta$

Therefore, the integral is

$I = \int \frac{d x}{x^{4} \sqrt{16 + x^{2}}} = \int \frac{4 {sec}^{2} θ d θ}{256 {tan}^{4} θ \cdot 4 sec θ}$ $I=int(dx)/(x^4sqrt(16+x^2))=int(4sec^2thetad theta)/(256tan^4theta*4sectheta)$

$= \frac{1}{256} \int \frac{sec θ d θ}{{tan}^{4} θ}$ $=1/256int(secthetad theta)/(tan^4theta)$

$tan θ = \frac{sin θ}{cos θ}$ $tantheta=sintheta/costheta$

$sec θ = \frac{1}{cos θ}$ $sectheta=1/costheta$

$I = \frac{1}{256} \int \frac{d θ}{cos θ \cdot \frac{{sin}^{4} θ}{{cos}^{4} θ}}$ $I=1/256int(d theta)/(costheta*sin^4theta/cos^4 theta)$

$= \frac{1}{256} \int \frac{{cos}^{3} θ d θ}{{sin}^{4} θ}$ $=1/256int(cos^3thetad theta)/(sin^4theta)$

$= \frac{1}{256} \int \frac{cos θ (1 - {sin}^{2} θ) d θ}{{sin}^{4} θ}$ $=1/256int(costheta(1-sin^2theta)d theta)/sin^4theta$

Let $v = sin θ$ $v=sintheta$ , $\Rightarrow$ $=>$ , $d v = cos θ d θ$ $dv=costheta d theta$

$I = \frac{1}{256} \int \frac{(1 - v^{2}) d v}{v^{4}}$ $I=1/256int((1-v^2)dv)/(v^4)$

$= \frac{1}{256} \int (\frac{1}{v^{4}} - \frac{1}{v^{2}}) d v$ $=1/256int(1/v^4-1/v^2)dv$

$= - \frac{1}{256} \cdot \frac{1}{3 v^{3}} + \frac{1}{256} \cdot (\frac{1}{v})$ $=-1/256*1/(3v^3)+1/256*(1/v)$

$= \frac{1}{256} \cdot \frac{1}{sin θ} - \frac{1}{768} \cdot \frac{1}{{sin}^{3} θ}$ $=1/256*1/sintheta-1/768*1/sin^3theta$

$= \frac{1}{256} \cdot \frac{1}{\frac{x}{\sqrt{16 + x^{2}}}} - \frac{1}{768} \cdot \frac{1}{\frac{x^{3}}{{(x^{2} + 16)}^{\frac{3}{2}}}} + C$ $=1/256*1/(x/sqrt(16+x^2))-1/768*1/(x^3/(x^2+16)^(3/2))+C$

$= \frac{\sqrt{16 + x^{2}}}{256 x} - \frac{{(x^{2} + 16)}^{\frac{3}{2}}}{768 x^{3}} + C$ $=sqrt(16+x^2)/(256x)-(x^2+16)^(3/2)/(768x^3)+C$

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How do you integrate $\int \frac{1}{x^{4} \sqrt{16 + x^{2}}}$ $int 1/(x^4sqrt(16+x^2))$ by trigonometric substitution?

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Explanation:

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How do you integrate $\int \frac{1}{x^{4} \sqrt{16 + x^{2}}}$ $int 1/(x^4sqrt(16+x^2))$ by trigonometric substitution?