Question

integral of (9+4*x^(-2/3))^(1/2)

Answer 1

Take the integral:

$$\int \sqrt {9 + \frac {4}{x ^ {\frac {4}{3}}}} d x$$

For the integrand $\sqrt{9 + \frac{4}{x^{\frac{4}{3}}}}$ , substitute $u = x^{\frac{-2}{3}}$ and $du = -\frac{2}{3x^{\frac{2}{3}}} dx$

$$\int \sqrt {9 + \frac {4}{x ^ {\frac {4}{3}}}} d x = - \frac {3}{2} \int \frac {\sqrt {4 u + 9}}{u ^ {\frac {2}{3}}} d u$$

For the integrand $\frac{\sqrt{4u + 9}}{u^{\frac{1}{2}}}$ , substitute $s = \sqrt{u}$ and $ds = \frac{1}{2\sqrt{u}} du$

integral $= -3\int \frac{\sqrt{4s^2 + 9}}{s^2} ds$

For the integrand $\frac{\sqrt{4s^2 + 9}}{s^2}$ , substitute $s = \frac{3\tan(p)}{2}$ and $ds = \frac{3}{2} \sec^2(p) dp$ .

Then $\sqrt{4s^2 + 9} = \sqrt{9\tan^2(p) + 9} = 3\sec(p)$ and $p = \tan^{-1}\left(\frac{2\pi}{3}\right)$

integral $= -\frac{27}{2}\int \frac{16}{81} cot(p)csc^3 (p)dp = -\frac{8}{3}\int cot(p)csc^3 (p)dp$

substitute $w = csc(p)$ and get

integral $= \frac{8\int w^2dw}{3} = \frac{8csc^3(p)}{9} +constant = = \frac{(4s^2 + 9)^{\frac{3}{2}}}{9s^2} +constant =$

$$= \frac {1}{9} x \left(\frac {4}{x ^ {\frac {4}{3}}} + 9\right) ^ {\frac {3}{2}} + c o n s t a n t$$