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Question Number 52233 by de best last updated on 04/Jan/19

please can you help me with this   caculus: ∫(1/(cos^2 x)) dx

$${please}\:{can}\:{you}\:{help}\:{me}\:{with}\:{this}\: \\ $$$${caculus}:\:\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}\:{dx} \\ $$

Answered by MJS last updated on 04/Jan/19

this is a standard integral  f(x)=tan x  f′(x)=(1/(cos^2  x))  ⇒ ∫(dx/(cos^2  x))=tan x +C

$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{standard}\:\mathrm{integral} \\ $$$${f}\left({x}\right)=\mathrm{tan}\:{x} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\Rightarrow\:\int\frac{{dx}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\mathrm{tan}\:{x}\:+{C} \\ $$

Answered by peter frank last updated on 04/Jan/19

∫sec^2 xdx=tan x+G

$$\int{sec}^{\mathrm{2}} {xdx}=\mathrm{tan}\:{x}+{G} \\ $$

Answered by peter frank last updated on 04/Jan/19

∫sec^2 xdx=tan x+G

$$\int{sec}^{\mathrm{2}} {xdx}=\mathrm{tan}\:{x}+{G} \\ $$

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