Integrate $\int \frac{x^{3}}{\sqrt{x^{2} + 4}}$ $intx^3/sqrt(x^2+4)$ using trig substitution?

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Answer 1 · 2017-10-11T06:43:38

See the explanatiom below

You have to change as follows

$I = 8 (\frac{1}{3} u^{3} - u)$ $I=8(1/3u^3-u)$

$I = \frac{8}{3} ({sec}^{3} θ - 3 sec θ)$ $I=8/3(sec^3theta-3sectheta)$

$= \frac{8}{3} ⎛ ⎝ {(\frac{x^{2} + 1}{2})}^{\frac{3}{2}} - 3 sec (arctan (\frac{x}{2})) + C$ $=8/3(((x^2+1)/2)^(3/2)-3sec(arctan(x/2))+C$

It's easier without trigonometric substitution

Let $u = x^{2} + 4$ $u=x^2+4$ , $\Rightarrow$ $=>$ , $d u = 2 x d x$ $du=2xdx$

$I = \frac{1}{2} \int \frac{(u - 4) d u}{\sqrt{u}}$ $I=1/2int((u-4)du)/sqrtu$

$= \frac{1}{2} \int \sqrt{u} d u - \int \frac{4}{\sqrt{u}} d u$ $=1/2intsqrtudu-int4/sqrtudu$

$= (\frac{u^{\frac{3}{2}}}{3} - 4 \sqrt{u})$ $=(u^(3/2)/3-4sqrtu)$

$= \frac{1}{3} {(x^{2} + 4)}^{\frac{3}{2}} - 4 \sqrt{x^{2} + 4}$ $=1/3(x^2+4)^(3/2)-4sqrt(x^2+4)$

$= \frac{(x^{2} - 8)}{3} \sqrt{x^{2} + 4} + C$ $=((x^2-8))/3sqrt(x^2+4)+C$

Integrate ∫x3√x2+4intx^3/sqrt(x^2+4) using trig substitution?