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

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

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Answer 1 · 2016-11-09T16:59:55

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Assuming you meant: $\int \frac{d x}{\sqrt{16 + x^{2}}}$ $intdx/sqrt(16+x^2)$

Let $x = 4 tan θ$ $x=4tantheta$ . This implies that $d x = 4 {sec}^{2} θ . d θ$ $dx=4sec^2thetacolor(white).d theta$ . Substituting these in yields:

$= \int \frac{4 {sec}^{2} θ . d θ}{\sqrt{16 + 16 {tan}^{2} θ}} = \frac{4}{\sqrt{16}} \int \frac{{sec}^{2} θ}{\sqrt{1 + {tan}^{2} θ}} d θ$ $=int(4sec^2thetacolor(white).d theta)/sqrt(16+16tan^2theta)=4/sqrt16intsec^2theta/sqrt(1+tan^2theta)d theta$

Recall that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ :

$= \int \frac{{sec}^{2} θ}{sec θ} d θ = \int sec θ . d θ = ln | tan θ + sec θ | + C$ $=intsec^2theta/secthetad theta=intsecthetacolor(white).d theta=lnabs(tantheta+sectheta)+C$

Rewriting in terms of tangent since our substitution is $tan θ = \frac{x}{4}$ $tantheta=x/4$ :

$= ln ∣ ∣ tan θ + \sqrt{1 + {tan}^{2} θ} ∣ ∣ + C = ln ∣ ∣ ∣ \frac{x}{4} + \sqrt{1 + \frac{x^{2}}{16}} ∣ ∣ ∣ + C$ $=lnabs(tantheta+sqrt(1+tan^2theta))+C=lnabs(x/4+sqrt(1+x^2/16))+C$

$= ln ∣ ∣ ∣ \frac{x}{4} + \frac{1}{4} \sqrt{16 + x^{2}} ∣ ∣ ∣ + C$ $=lnabs(x/4+1/4sqrt(16+x^2))+C$

Factoring the $\frac{1}{4}$ $1/4$ and moving it out of the integral as the constant $ln (\frac{1}{4})$ $ln(1/4)$ through the $log (A B) = log (A) + log (B)$ $log(AB)=log(A)+log(B)$ rule, we see it combines with the constant of integration:

$= ln ∣ ∣ x + \sqrt{16 + x^{2}} ∣ ∣ + C$ $=lnabs(x+sqrt(16+x^2))+C$

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Assuming you meant: $\int \frac{d x}{{(\sqrt{16 + x^{2}})}^{2}}$ $intdx/(sqrt(16+x^2))^2$

The square root and the exponent cancel, leaving just:

$= \int \frac{d x}{16 + x^{2}}$ $=intdx/(16+x^2)$

Now use the same substitution as before, $x = 4 tan θ$ $x=4tantheta$ such that $d x = 4 {sec}^{2} θ . d θ$ $dx=4sec^2thetacolor(white).d theta$ .

$= \int \frac{4 {sec}^{2} θ . d θ}{16 + 16 {tan}^{2} θ} = \frac{1}{4} \int \frac{{sec}^{2} θ}{1 + {tan}^{2} θ} d θ = \frac{1}{4} \int d θ$ $=int(4sec^2thetacolor(white).d theta)/(16+16tan^2theta)=1/4intsec^2theta/(1+tan^2theta)d theta=1/4intd theta$

$= \frac{1}{4} θ + C$ $=1/4theta+C$

From the substitution $x = 4 tan θ$ $x=4tantheta$ solving for $θ$ $theta$ yields:

$= \frac{1}{4} arctan (\frac{x}{4}) + C$ $=1/4arctan(x/4)+C$

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

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