Question Number 45795 by maxmathsup by imad last updated on 16/Oct/18
$${find}\:\int\:\frac{{dx}}{{cosx}\:{sin}^{\mathrm{2}} {x}} \\ $$
Answered by MJS last updated on 17/Oct/18
![(1/(cos x sin^2 x))=(1+((cos^2 x)/(sin^2 x)))(1/(cos x)) = =(1/(cos x))+((cos x)/(sin^2 x)) ∫(dx/(cos x))=∫sec x dx=ln ∣tan x +sec x∣ =ln ∣((1+sin x)/(cos x))∣ ∫((cos x)/(sin^2 x))dx= [t=sin x → dx=(dt/(cos x))] =∫(dt/t^2 )=−(1/t)=−(1/(sin x)) ∫(dx/(cos x sin^2 x))=ln ∣((1+sin x)/(cos x))∣ −(1/(sin x))+C](https://www.tinkutara.com/question/Q45798.png)
$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}=\left(\mathrm{1}+\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}\right)\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$\int\frac{{dx}}{\mathrm{cos}\:{x}}=\int\mathrm{sec}\:{x}\:{dx}=\mathrm{ln}\:\mid\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\mid\:=\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\mid \\ $$$$\int\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}} \\ $$$$\int\frac{{dx}}{\mathrm{cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}=\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\mid\:−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+{C} \\ $$
Commented by math khazana by abdo last updated on 17/Oct/18
$${thank}\:{you}\:{sir}\:{MJS}. \\ $$
Commented by MJS last updated on 17/Oct/18
![btw (1/(cos^2 x sin x))=(1+((sin^2 x)/(cos^2 x)))(1/(sin x))= =(1/(sin x))+((sin x)/(cos^2 x)) ∫(dx/(sin x))=∫csc x dx=−ln ∣cot x +csc x∣ =−ln ∣((1+cos x)/(sin x))∣ ∫((sin x)/(cos^2 x))dx= [t=cos x → dx=−(dt/(sin x))] =−∫(dt/t^2 )=(1/t)=(1/(cos x))](https://www.tinkutara.com/question/Q45800.png)
$$\mathrm{btw} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}\:{x}}=\left(\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\right)\frac{\mathrm{1}}{\mathrm{sin}\:{x}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\int\frac{{dx}}{\mathrm{sin}\:{x}}=\int\mathrm{csc}\:{x}\:{dx}=−\mathrm{ln}\:\mid\mathrm{cot}\:{x}\:+\mathrm{csc}\:{x}\mid\:=−\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\mid \\ $$$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{sin}\:{x}}\right] \\ $$$$=−\int\frac{{dt}}{{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{{t}}=\frac{\mathrm{1}}{\mathrm{cos}\:{x}} \\ $$
Commented by MJS last updated on 17/Oct/18
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}.\:\mathrm{I}\:\mathrm{just}\:\mathrm{fixed}\:\mathrm{a}\:\mathrm{typo}… \\ $$