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Question Number 19457 by NEC last updated on 11/Aug/17

prove that ∫tan^2 xdx =tan x−x

$${prove}\:{that}\:\int\mathrm{tan}\:^{\mathrm{2}} {xdx} \\ $$$$ \\ $$$$=\mathrm{tan}\:{x}−{x} \\ $$

Answered by Joel577 last updated on 11/Aug/17

∫ tan^2 x dx = ∫ (sec^2 x − 1) dx = tan x − x + C

$$\int\:\mathrm{tan}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$=\:\int\:\left(\mathrm{sec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\right)\:{dx} \\ $$$$=\:\mathrm{tan}\:{x}\:−\:{x}\:+\:{C} \\ $$

Commented by NEC last updated on 11/Aug/17

how is the integral of sec^2 x=tanx. prove it please.

$${how}\:{is}\:{the}\:{integral}\:{of}\:{sec}^{\mathrm{2}} {x}={tanx}. \\ $$$${prove}\:{it}\:{please}. \\ $$

Commented by Tinkutara last updated on 11/Aug/17

Because if y = tan x then (dy/dx) = sec^2 x So ∫sec^2 x dx = tan x + C

$$\mathrm{Because}\:\mathrm{if}\:{y}\:=\:\mathrm{tan}\:{x}\:\mathrm{then}\:\frac{{dy}}{{dx}}\:=\:\mathrm{sec}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{So}\:\int\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx}\:=\:\mathrm{tan}\:{x}\:+\:{C} \\ $$

Commented by Joel577 last updated on 12/Aug/17

Let y = tan x = ((sin x)/(cos x)) (dy/dx) = (((d/dx)(sin x) . cos x − (d/dx)(cos x) . sin x )/(cos^2 x)) = ((cos^2 x + sin^2 )/(cos^2 x)) = (1/(cos^2 x)) = sec^2 x Hence, ∫ sec^2 x dx = tan x + C

$$\mathrm{Let}\:{y}\:=\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\frac{{d}}{{dx}}\left(\mathrm{sin}\:{x}\right)\:.\:\mathrm{cos}\:{x}\:−\:\frac{{d}}{{dx}}\left(\mathrm{cos}\:{x}\right)\:.\:\mathrm{sin}\:{x}\:}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\:\mathrm{sin}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\:=\:\mathrm{sec}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{Hence},\:\int\:\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx}\:=\:\mathrm{tan}\:\:{x}\:+\:{C} \\ $$$$ \\ $$

Commented by ajfour last updated on 12/Aug/17

Commented by ajfour last updated on 12/Aug/17

the greater angle at bottom left corner is x+dx and not just x .

$$\mathrm{the}\:\mathrm{greater}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{bottom}\:\mathrm{left} \\ $$$$\mathrm{corner}\:\mathrm{is}\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{dx}}\:\mathrm{and}\:\mathrm{not}\:\mathrm{just}\:\boldsymbol{\mathrm{x}}\:. \\ $$

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