definite integral, from 0 to 1, of x^3 / (4 + x^2)^(1/2)

[clw89]

Hey I need help with the following integral:

the integral from 0 to 1 of: x[sup:24y84zxd]3[/sup:24y84zxd]/(4+x[sup:24y84zxd]2[/sup:24y84zxd])[sup:24y84zxd]1/2[/sup:24y84zxd]

thanks!!

 

[soroban]

Hello, clw89!

$\displaystyle \text{Evaluate: }\:\int^1_0\frac{x^3}{\sqrt{4+x^2}}\,dx$

I would try Trig Substitution . . .


\(\displaystyle \text{Let: }\,x \:=\:\2tan\theta\quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta\,d\theta\)
. . $\displaystyle \text{and: }\:\sqrt{4+x^2} \:=\:2\sec\theta$

$\displaystyle \text{Substitute: }\;\int\frac{8\tan^3\!\theta}{2\sec\theta}\,(2\sec^2\!\theta\,d\theta) \;=\;8\int\sec\theta\tan^3\!\theta\,d\theta$

. . $\displaystyle =\;8\int\tan^2\!\theta(\sec\theta\tan\theta\,d\theta) \;=\;8\int(\sec^2\!\theta-1)(\sec\theta\tan\theta\,d\theta)$


. . $\displaystyle \text{Now let }\,u \:=\:\sec\theta$ . . .

 

[skeeter]

$\displaystyle \int_0^1 \frac{x^3}{\sqrt{4+x^2}} dx$

$\displaystyle u = 4 + x^2$

$\displaystyle x^2 = u - 4$

$\displaystyle du = 2x \, dx$

$\displaystyle \frac{1}{2} \int_0^1 \frac{x^2 \cdot 2x}{\sqrt{4+x^2}} dx$

$\displaystyle \frac{1}{2} \int_4^5 \frac{u-4}{\sqrt{u}} \, du$

\(\displaystyle \frac{1}{2} \int_4^5 u^{\frac{1}{2}} - 4u^{-\frac{1}{2} \, du\)