If-F-x-0-arctanx-t-4-1-t-4-1-dt-find-F-x-

Question Number 78468 by Dah Solu Tion last updated on 17/Jan/20

I f F (x) = \int_{0}^{a r c t a n x} \sqrt{\frac{t^{4} - 1}{t^{4} + 1} d t,} f i n d F^{'} (x) .

Commented by mathmax by abdo last updated on 17/Jan/20

w e h a v e F (x) = \int_{0}^{a r c t a n (x)} \sqrt{\frac{t^{4} - 1}{t^{4} + 1}} d t \Rightarrow

F (x) = \int_{0}^{a r c t a n (x)} \sqrt{\frac{t^{4} + 1 - 2}{t^{4} + 1}} d t = \int_{0}^{a r c t a n x} {(1 - \frac{2}{t^{4} + 1})}^{\frac{1}{2}} d t

\frac{d F}{d x} (x) = \frac{1}{1 + {t a n}^{2} x} {(1 - \frac{2}{1 + {t a n}^{4} x})}^{\frac{1}{2}} = {c o s}^{2} x {(1 - \frac{2}{(1 + {t a n}^{2} x) - 2 t a n x})}^{\frac{1}{2}}

= {c o s}^{2} x {(1 - \frac{2}{{c o s}^{4} x - 2 t a n x})}^{\frac{1}{2}}