Using trigonometric substitution, how do you integrate integral of $x^{3} {(x^{2} + 4)}^{\frac{1}{2}}$ $x^3 (x^2+4)^(1/2)$ ?

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Using trigonometric substitution, how do you integrate integral of $x^{3} {(x^{2} + 4)}^{\frac{1}{2}}$ $x^3 (x^2+4)^(1/2)$ ?

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Answer 1 · 2018-07-21T19:18:04

$\int x^{3} {(x^{2} + 4)}^{\frac{1}{2}} d x = \frac{1}{5} {(4 + x^{2})}^{2} \sqrt{4 + x^{2}} - \frac{4}{3} (4 + x^{2}) \sqrt{4 + x^{2}}$ $intx^3(x^2+4)^(1/2)dx=1/5(4+x^2)^2sqrt(4+x^2)-4/3(4+x^2)sqrt(4+x^2)$

See explanations below.

Explanation:

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

let $x = 2 tan (θ)$ $x=2tan(theta)$

$d x = 2 {sec (θ)}^{2} d θ$ $dx=2sec(theta)^2d theta$

So :

$\int x^{3} {(x^{2} + 4)}^{\frac{1}{2}} d x = \int 16 {tan (θ)}^{3} \sqrt{(4 {tan (θ)}^{2} + 4)} {sec (θ)}^{2} d θ$ $intx^3(x^2+4)^(1/2)dx=int16tan(theta)^3sqrt((4tan(theta)^2+4))sec(theta)^2d theta$

$= 16 \int {tan (θ)}^{3} \sqrt{4 ({tan (θ)}^{2} + 1)} {sec (θ)}^{2} d θ$ $=16inttan(theta)^3sqrt(4(tan(theta)^2+1))sec(theta)^2d theta$

$=32inttan(theta)^3cancel(sqrt(tan(theta)^2+1))^(color(red)(=sec(theta)))sec(theta)^2d theta$

$=32inttan(theta)^3sec(theta)^3d theta$

$=32int(sin(theta)^3)/(cos(theta)^6)d theta$

$=-32int(-sin(theta)sin(theta)^2)/(cos(theta)^6)d theta$

$=-32int(-sin(theta)(1-cos(theta)^2))/(cos(theta)^6)d theta$

let $u=cos(theta)$
$du=-sin(theta)d theta$

So : $32inttan(theta)^3sec(theta)^3d theta=-32int(1-u^2)/(u^6)du$

$=32intu^(-4)du-32intu^(-6)du$

$=32/5u^(-5)-32/3u^(-3)$

$=32/5cos(theta)^(-5)-32/3cos(theta)^-3$

Finally, because $theta=arctan(x/2)$ and $cos(arctan(x/2))=2/sqrt(4+x^2)$ ,

$intx^3(x^2+4)^(1/2)dx=1/5(4+x^2)^2sqrt(4+x^2)-4/3(4+x^2)sqrt(4+x^2)$

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Using trigonometric substitution, how do you integrate integral of $x^{3} {(x^{2} + 4)}^{\frac{1}{2}}$ $x^3 (x^2+4)^(1/2)$ ?

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