I-sec-x-1-csc-x-dx-

Question Number 141847 by iloveisrael last updated on 24/May/21

$$\:\mathcal{I}\:=\:\int\:\frac{\mathrm{sec}\:{x}}{\mathrm{1}+\mathrm{csc}\:{x}}\:{dx}\: \\ $$

Answered by MJS_new last updated on 24/May/21

∫((sec x)/(1+csc x))= [t=tan (x/2) → dx=2cos^2 (x/2) dt] =−4∫(t/((t−1)(t+1)^3 ))dt= =−(1/2)∫(dt/(t−1))−2∫(dt/((t+1)^3 ))+∫(dt/((t+1)^2 ))+(1/2)∫(dt/(t+1))= =−(1/2)ln (t−1) +(1/((t+1)^2 ))−(1/(t+1))+(1/2)ln (t+1) = =−(t/((t+1)^2 ))+(1/2)ln ((t+1)/(t−1)) =... =−((sin x)/(2(1+sin x)))+(1/2)ln ∣((cos x)/(1−sin x))∣ +C

$$\int\frac{\mathrm{sec}\:{x}}{\mathrm{1}+\mathrm{csc}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:{dt}\right] \\ $$$$=−\mathrm{4}\int\frac{{t}}{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}−\mathrm{1}}−\mathrm{2}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:= \\ $$$$=−\frac{{t}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\:=… \\ $$$$=−\frac{\mathrm{sin}\:{x}}{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\mid\:+{C} \\ $$

Answered by iloveisrael last updated on 24/May/21

I= ∫ ((1/(cos x))/(1+(1/(sin x)))) dx = ∫ ((sin x)/(cos xsin x+cos x)) dx I= ∫ ((sin x)/(cos x(sin x+1))) dx I=∫ ((sin x(sin x−1))/(−cos^3 x)) dx I=−∫ ((sin^2 x−sin x)/(cos^3 x)) dx I=−{∫ sec^3 x dx −∫ sec x dx }− ∫ ((d(cos x))/(cos^3 x)) I=ln ∣sec x+tan x∣+ (1/2)sec^2 x − ∫ sec^3 x dx

$${I}=\:\int\:\frac{\frac{\mathrm{1}}{\mathrm{cos}\:{x}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:{x}}}\:{dx}\:=\:\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx} \\ $$$${I}=\:\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\left(\mathrm{sin}\:{x}+\mathrm{1}\right)}\:{dx} \\ $$$${I}=\int\:\frac{\mathrm{sin}\:{x}\left(\mathrm{sin}\:{x}−\mathrm{1}\right)}{−\mathrm{cos}\:^{\mathrm{3}} {x}}\:{dx} \\ $$$${I}=−\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:^{\mathrm{3}} {x}}\:{dx} \\ $$$${I}=−\left\{\int\:\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}\:−\int\:\mathrm{sec}\:{x}\:{dx}\:\right\}− \\ $$$$\:\:\:\:\:\:\:\int\:\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\mathrm{cos}\:^{\mathrm{3}} {x}}\: \\ $$$${I}=\mathrm{ln}\:\mid\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\mid+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} {x}\:− \\ $$$$\:\:\:\:\:\:\int\:\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}\: \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply