Menu Close

cosec-2-xdx-




Question Number 20166 by tammi last updated on 23/Aug/17
∫cosec^2 xdx
$$\int\mathrm{cosec}\:^{\mathrm{2}} {xdx} \\ $$
Answered by sma3l2996 last updated on 23/Aug/17
t=cosecx⇒dt=−((cosec^2 x)/(secx))dx  cosec^2 xdx=−((tdt)/( (√(t^2 −1))))  ∫cosec^2 xdx=−∫((tdt)/( (√(t^2 −1))))=−∫((d(t^2 −1))/(2(√(t^2 −1))))  =−(√(t^2 −1))+C=−(√(cosec^2 x−1))+C  ∫cosec^2 xdx=−cotx+C
$${t}={cosecx}\Rightarrow{dt}=−\frac{{cosec}^{\mathrm{2}} {x}}{{secx}}{dx} \\ $$$${cosec}^{\mathrm{2}} {xdx}=−\frac{{tdt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\int{cosec}^{\mathrm{2}} {xdx}=−\int\frac{{tdt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}=−\int\frac{{d}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=−\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}+{C}=−\sqrt{{cosec}^{\mathrm{2}} {x}−\mathrm{1}}+{C} \\ $$$$\int{cosec}^{\mathrm{2}} {xdx}=−{cotx}+{C} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *