Question Number 19457 by NEC last updated on 11/Aug/17
$${prove}\:{that}\:\int\mathrm{tan}\:^{\mathrm{2}} {xdx} \\ $$$$ \\ $$$$=\mathrm{tan}\:{x}−{x} \\ $$
Answered by Joel577 last updated on 11/Aug/17
$$\int\:\mathrm{tan}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$=\:\int\:\left(\mathrm{sec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\right)\:{dx} \\ $$$$=\:\mathrm{tan}\:{x}\:−\:{x}\:+\:{C} \\ $$
Commented by NEC last updated on 11/Aug/17
$${how}\:{is}\:{the}\:{integral}\:{of}\:{sec}^{\mathrm{2}} {x}={tanx}. \\ $$$${prove}\:{it}\:{please}. \\ $$
Commented by Tinkutara last updated on 11/Aug/17
$$\mathrm{Because}\:\mathrm{if}\:{y}\:=\:\mathrm{tan}\:{x}\:\mathrm{then}\:\frac{{dy}}{{dx}}\:=\:\mathrm{sec}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{So}\:\int\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx}\:=\:\mathrm{tan}\:{x}\:+\:{C} \\ $$
Commented by Joel577 last updated on 12/Aug/17
$$\mathrm{Let}\:{y}\:=\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\frac{{d}}{{dx}}\left(\mathrm{sin}\:{x}\right)\:.\:\mathrm{cos}\:{x}\:−\:\frac{{d}}{{dx}}\left(\mathrm{cos}\:{x}\right)\:.\:\mathrm{sin}\:{x}\:}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\:\mathrm{sin}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\:=\:\mathrm{sec}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{Hence},\:\int\:\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx}\:=\:\mathrm{tan}\:\:{x}\:+\:{C} \\ $$$$ \\ $$
Commented by ajfour last updated on 12/Aug/17
Commented by ajfour last updated on 12/Aug/17
$$\mathrm{the}\:\mathrm{greater}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{bottom}\:\mathrm{left} \\ $$$$\mathrm{corner}\:\mathrm{is}\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{dx}}\:\mathrm{and}\:\mathrm{not}\:\mathrm{just}\:\boldsymbol{\mathrm{x}}\:. \\ $$