Question Number 145856 by mathdanisur last updated on 08/Jul/21
Commented by mr W last updated on 08/Jul/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(…\right){dv}={C}={constant} \\ $$$${f}\left({x}\right)=\frac{{d}}{{dx}}\left({C}\right)=\mathrm{0} \\ $$$${f}'\left({x}\right)=\mathrm{0} \\ $$$${f}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Commented by mathdanisur last updated on 09/Jul/21
$${thankyou}\:{Ser} \\ $$
Answered by mathmax by abdo last updated on 08/Jul/21
$$\mathrm{perhaps}\:\mathrm{the}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\int_{\mathrm{0}} ^{\mathrm{x}} \left(\mathrm{1}−\mathrm{tan}^{\mathrm{3}} \mathrm{v}\right)\mathrm{dv}… \\ $$