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

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Answer 1 · 2018-07-21T19:18:04

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

See explanations below.

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

let $x=2tan(theta)$

$dx=2sec(theta)^2d theta$

So :

$intx^3(x^2+4)^(1/2)dx=int16tan(theta)^3sqrt((4tan(theta)^2+4))sec(theta)^2d theta$

$=16inttan(theta)^3sqrt(4(tan(theta)^2+1))sec(theta)^2d theta$

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

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

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

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

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

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)^(1/2) x^3 (x^2+4)^(1/2) ?