How do you integrate $\int \frac{1}{{(x^{2} + 4)}^{\frac{3}{2}}}$ $int 1/(x^2+4)^(3/2)$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{{(x^{2} + 4)}^{\frac{3}{2}}}$ $int 1/(x^2+4)^(3/2)$ by trigonometric substitution?

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Answer 1 · 2018-04-01T17:48:14

$\int \frac{d x}{{(x^{2} + 4)}^{\frac{3}{2}}} = \frac{1}{4} \frac{x}{\sqrt{x^{2} + 4}} + C$ $intdx/(x^2+4)^(3/2)=1/4x/sqrt(x^2+4)+C$

Explanation:

Rewrite with the rational exponent converted to a root:

$\int \frac{d x}{\sqrt{{(x^{2} + 4)}^{3}}}$ $intdx/(sqrt((x^2+4)^3)$

Let $x = 2 tan θ$ $x=2tantheta$
$d x = 2 {sec}^{2} θ d θ$ $dx=2sec^2thetad theta$

Rewrite:

$\int \frac{2 {sec}^{2} θ}{\sqrt{{(4 {tan}^{2} θ + 4)}^{3}}} d θ$ $int(2sec^2theta)/(sqrt((4tan^2theta+4)^3))d theta$

$\int \frac{2 {sec}^{2} θ}{\sqrt{{(4 ({tan}^{2} θ + 1))}^{3}}} d θ$ $int(2sec^2theta)/(sqrt((4(tan^2theta+1))^3))d theta$

Recall the identity

${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ , and apply it:

$\int \frac{2 {sec}^{2} θ}{\sqrt{64 {({sec}^{2} θ)}^{3}}} d θ$ $int(2sec^2theta)/(sqrt(64(sec^2theta)^3))d theta$

$\frac{1}{4} \int \frac{{sec}^{2} θ}{\sqrt{{sec}^{6} θ}} d θ$ $1/4intsec^2theta/sqrt(sec^6theta)d theta$

$\frac{1}{4} \int \frac{{sec}^{2} θ}{{sec}^{3} θ} d θ$ $1/4intsec^2theta/sec^3thetad theta$

$\frac{1}{4} \int cos θ d θ = \frac{1}{4} sin θ + C$ $1/4intcosthetad theta=1/4sintheta+C$

We want to rewrite in terms of $x .$ $x.$ Recalling that $x = 2 tan θ, tan θ = \frac{x}{2} .$ $x=2tantheta, tantheta=x/2.$ If $tan θ = \frac{x}{2}, sin θ = \frac{x}{\sqrt{x^{2} + 4}}$ $tantheta=x/2, sintheta=x/sqrt(x^2+4)$ .

This could be deduced by drawing a right triangle with the sides opposite and adjacent to angle $θ$ $theta$ being labeled $x, 2$ $x, 2$ (respectively), making the hypotenuse $\sqrt{x^{2} + 4}$ $sqrt(x^2+4)$ , and $sin θ = \frac{o p p o s i t e}{h y p o t e n u s e} = \frac{x}{\sqrt{x^{2} + 4}}$ $sintheta=(opposite)/(hypoten use)=x/sqrt(x^2+4)$

So,

$\int \frac{d x}{{(x^{2} + 4)}^{\frac{3}{2}}} = \frac{1}{4} \frac{x}{\sqrt{x^{2} + 4}} + C$ $intdx/(x^2+4)^(3/2)=1/4x/sqrt(x^2+4)+C$

Answer 2 · 2018-04-01T18:22:32

$I = \frac{1}{4} \cdot \frac{x}{\sqrt{x^{2} + 4}} + c$ $I=1/4*x/sqrt(x^2+4)+c$

Explanation:

Here,

$I = \int \frac{1}{{(x^{2} + 4)}^{\frac{3}{2}}} d x$ $I=int1/(x^2+4)^(3/2)dx$

Let, $x = 2 tan u \Rightarrow d x = 2 {sec}^{2} u d u$ $x=2tanu=>dx=2sec^2udu$

$and x^{2} + 4 = 4 {tan}^{2} u + 4 = 4 ({tan}^{2} u + 1) = 4 {sec}^{2} u$ $andx^2+4=4tan^2u+4=4(tan^2u+1)=4sec^2u$

So,

$I = \int \frac{2 {sec}^{2} u}{{(4 {sec}^{2} u)}^{\frac{3}{2}}} d u = \int \frac{2 {sec}^{2} u}{{(2 sec u)}^{3}} d u$ $I=int(2sec^2u)/((4sec^2u)^(3/2))du=int(2sec^2u)/(2secu)^3du$

$\Rightarrow I = \int \frac{2 {sec}^{2} u}{8 {sec}^{3} u} d u$ $=>I=int(2sec^2u)/(8sec^3u)du$

$\Rightarrow I = \int \frac{1}{4 sec u} d u$ $=>I=int1/(4secu)du$

$\Rightarrow I = \frac{1}{4} \int cos d u$ $=>I=1/4intcosdu$

$\Rightarrow I = \frac{1}{4} sin u + c$ $=>I=1/4sinu+c$

$\Rightarrow I = \frac{1}{2} (\frac{sin u}{cos u}) cos u + c$ $=>I=1/2(sinu/cosu)cosu+c$

$= \frac{1}{4} \frac{tan u}{sec u} + c$ $=1/4tanu/secu+c$

$= \frac{1}{4} \frac{tan u}{\sqrt{{tan}^{2} u + 1}} + c$ $=1/4tanu/sqrt(tan^2u+1)+c$

$= \frac{1}{4} \frac{\frac{x}{2}}{\sqrt{(\frac{x^{2}}{4}) + 1}} + c$ $=1/4(x/2)/sqrt((x^2/4)+1)+c$

$= \frac{1}{4} \cdot \frac{x}{\sqrt{x^{2} + 4}} + c$ $=1/4*x/sqrt(x^2+4)+c$

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How do you integrate $\int \frac{1}{{(x^{2} + 4)}^{\frac{3}{2}}}$ $int 1/(x^2+4)^(3/2)$ by trigonometric substitution?

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