A calculus problem by Russel Ryan Floresca

Length of curve

$\displaystyle =\int \Big( \sqrt{(dx)^2 +(dy)^2} \Big)$

$\displaystyle =\int_{0}^9 \Bigg( \sqrt{\Big( dx \Big)^2 + \Big( \frac{4}{5} \times \frac{5}{4} \times x^{\frac{1}{4}} dx \Big) ^2} \Bigg)$

$\displaystyle =\int_{0}^9 \Big( \sqrt{1+\sqrt{x}} \Big) dx$

Substituting $\displaystyle x = \tan^4\theta$ :

$\displaystyle =\int_{0}^\frac{\pi}{3} \sqrt{1+\tan^2\theta}\cdot 4\tan^3\theta\sec^2\theta d\theta$

$\displaystyle =4 \int_{0}^\frac{\pi}{3} (\sec^2\theta - 1)\sec^2\theta \cdot \sec\theta \tan\theta d\theta$

$\displaystyle =4\int_{0}^\frac{\pi}{3} (\sec^4\theta - \sec^2\theta) \cdot \sec\theta \tan\theta d\theta$

$\displaystyle = 4\Bigg( \Big( \frac{ \sec^5( \frac{ \pi}{3}) - \sec^5(0)}{5} \Big) - \Big( \frac{ \sec^3( \frac{ \pi}{3}) - \sec^3(0)}{3} \Big) \Bigg)$

$\displaystyle = \frac{232}{15}$

$\displaystyle \approx \boxed{15.467}$

@Calvin Lin sir, can you edit my problem sir? I can't use LATEX at the moment. Sorry for the quality and the size. Thanks in advance!