sec-x-tan-2-x-dx-

Question Number 12768 by Joel577 last updated on 01/May/17

$$\int\:\frac{\mathrm{sec}\:{x}}{\mathrm{tan}^{\mathrm{2}} \:{x}}\:{dx} \\ $$

Answered by prakash jain last updated on 01/May/17

∫ ((sec x∙cos^2 x)/(sin^2 x)) dx =∫ ((cos x)/(sin^2 x))dx sin x=u⇒cos xdx=du =∫(du/u^2 )=−(1/u)+C=−(1/(sin x))+C

$$\int\:\frac{\mathrm{sec}\:{x}\centerdot\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}^{\mathrm{2}} {x}}\:\mathrm{d}{x} \\ $$$$=\int\:\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{sin}\:{x}={u}\Rightarrow\mathrm{cos}\:{xdx}={du} \\ $$$$=\int\frac{{du}}{{u}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{u}}+{C}=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+{C} \\ $$

Commented by Joel577 last updated on 01/May/17