Question Number 182648 by cortano1 last updated on 12/Dec/22
$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$
Answered by Acem last updated on 12/Dec/22
![tan x − sec x= t ..... m sec x (sec x− tan x) dx= dt sec x dx= ((−dt)/t) ... i ∗ sec^2 x − tan^2 x= 1 ⇔ (sec x+ tan x) (sec x− tan x)=1 ⇒ sec x + tan x= ((−1)/t) ∗ & sec x − tan x= −t ∗∗^( From m) Sum stars: sec x= −(1/2)((1/t) + t) ...ii i, ii in integ. : I= ∫ ((sec x sec x dx)/(tan x− sec x))= (1/2)∫ ((((1/t) + t))/t^( 2) ) dt I= (1/2) ∫ ((1/t) + (1/t^3 )) dt, from m { ((m=0 : t= −1)),((m= (π/4) : t= 1−(√2) )) :} a= (1/2) ∫_(−1) ^( 1−(√2)) ((1/t) + (1/t^3 )) dt a= (1/2) [ln ∣t∣ − (1/(2 t^2 ))]∣_( −1) ^(1−(√2)) ≈ − 1. 648](Q182665.png)
$$\:\mathrm{tan}\:{x}\:−\:\mathrm{sec}\:{x}=\:{t}\:\:.....\:{m} \\ $$$$\:\mathrm{sec}\:{x}\:\left(\mathrm{sec}\:{x}−\:\mathrm{tan}\:{x}\right)\:{dx}=\:{dt} \\ $$$$\:\mathrm{sec}\:{x}\:{dx}=\:\frac{−{dt}}{{t}}\:\:...\:{i} \\ $$$$\:\ast\:\mathrm{sec}^{\mathrm{2}} \:{x}\:−\:\mathrm{tan}^{\mathrm{2}} \:{x}=\:\mathrm{1}\:\Leftrightarrow\:\left(\mathrm{sec}\:{x}+\:\mathrm{tan}\:{x}\right)\:\left(\mathrm{sec}\:{x}−\:\mathrm{tan}\:{x}\right)=\mathrm{1} \\ $$$$\:\Rightarrow\:\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}=\:\frac{−\mathrm{1}}{{t}}\:\:\:\ast \\ $$$$\:\&\:\:\mathrm{sec}\:{x}\:−\:\mathrm{tan}\:{x}=\:−{t}\:\:\:\ast\ast^{\:{From}\:{m}} \\ $$$$\:{Sum}\:{stars}:\:\mathrm{sec}\:{x}=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{t}}\:+\:{t}\right)\:...{ii}\:\:\:\: \\ $$$$\:{i},\:{ii}\:{in}\:{integ}.\::\:{I}=\:\int\:\frac{\mathrm{sec}\:{x}\:\mathrm{sec}\:{x}\:{dx}}{\mathrm{tan}\:{x}−\:\mathrm{sec}\:{x}}=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\left(\frac{\mathrm{1}}{{t}}\:+\:{t}\right)}{{t}^{\:\mathrm{2}} }\:{dt} \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\frac{\mathrm{1}}{{t}}\:+\:\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right)\:{dt},\:\:{from}\:{m\begin{cases}{{m}=\mathrm{0}\:\:\::\:{t}=\:−\mathrm{1}}\\{{m}=\:\frac{\pi}{\mathrm{4}}\::\:{t}=\:\mathrm{1}−\sqrt{\mathrm{2}}\:\:}\end{cases}} \\ $$$$\:{a}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{1}} ^{\:\mathrm{1}−\sqrt{\mathrm{2}}} \:\left(\frac{\mathrm{1}}{{t}}\:+\:\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right)\:{dt} \\ $$$$\:{a}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\mathrm{ln}\:\mid{t}\mid\:−\:\frac{\mathrm{1}}{\mathrm{2}\:{t}^{\mathrm{2}} }\right]\mid_{\:−\mathrm{1}} ^{\mathrm{1}−\sqrt{\mathrm{2}}} \:\:\:\approx\:−\:\mathrm{1}.\:\mathrm{648} \\ $$$$ \\ $$
Answered by FelipeLz last updated on 12/Dec/22
![tan(x) = u → du = sec^2 (x)dx x = 0 → u = 0 x = (π/4) → u = 1 I =∫_0 ^(π/4) ((sec^2 (x))/(tan(x)−sec(x)))dx = ∫_0 ^1 (1/(u−(√(u^2 +1))))du = ∫_0 ^1 (1/(u−(√(u^2 +1))))×((u+(√(u^2 +1)))/(u+(√(u^2 +1))))du = ∫_0 ^1 ((u+(√(u^2 +1)))/(u^2 −u^2 −1))du I = −∫_0 ^1 udu−∫_0 ^1 (√(u^2 +1))du I = −∫_0 ^(π/4) tan(x)sec^2 (x)dx−∫_0 ^(π/4) sec^3 (x)dx ∫tan(x)sec^2 (x)dx = sec^2 (x)−∫tan(x)sec^2 (x)dx 2∫tan(x)sec^2 (x)dx = sec^2 (x) ∫tan(x)sec^2 (x)dx = (1/2)sec^2 (x) ∫sec^3 (x)dx = sec(x)tan(x)−∫sec(x)tan^2 (x)dx ∫sec^3 (x)dx = sec(x)tan(x)−∫sec^3 (x)dx+∫sec(x)dx 2∫sec^3 (x)dx = sec(x)tan(x)+ln∣sec(x)+tan(x)∣ ∫sec^3 (x)dx = (1/2)sec(x)tan(x)+(1/2)ln∣sec(x)+tan(x)∣ I = −(1/2)[sec^2 (x)]_0 ^(π/4) −(1/2)[sec(x)tan(x)+ln∣sec(x)+tan(x)∣]_0 ^(π/4) I = −(1/2)[((√2))^2 −1^2 ]−(1/2)[(√2)×1+ln∣(√2)+1∣−1×0−ln∣1+0∣] I = −(1/2)[1+(√2)+ln(1+(√2))]](Q182670.png)
$$\:\:\:\:\:\mathrm{tan}\left({x}\right)\:=\:{u}\:\rightarrow\:{du}\:=\:\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$\:\:\:\:\:{x}\:=\:\mathrm{0}\:\rightarrow\:{u}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:{x}\:=\:\frac{\pi}{\mathrm{4}}\:\rightarrow\:{u}\:=\:\mathrm{1} \\ $$$${I}\:=\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{sec}^{\mathrm{2}} \left({x}\right)}{\mathrm{tan}\left({x}\right)−\mathrm{sec}\left({x}\right)}{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{u}−\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{du}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{u}−\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}×\frac{{u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{{u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{du}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{{u}^{\mathrm{2}} −{u}^{\mathrm{2}} −\mathrm{1}}{du}\:\: \\ $$$${I}\:=\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{udu}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}{du}\: \\ $$$${I}\:=\:−\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{tan}\left({x}\right)\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx}−\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{sec}^{\mathrm{3}} \left({x}\right){dx} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\int\mathrm{tan}\left({x}\right)\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx}\:=\:\mathrm{sec}^{\mathrm{2}} \left({x}\right)−\int\mathrm{tan}\left({x}\right)\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$\:\:\:\:\:\mathrm{2}\int\mathrm{tan}\left({x}\right)\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx}\:=\:\mathrm{sec}^{\mathrm{2}} \left({x}\right) \\ $$$$\:\:\:\:\:\int\mathrm{tan}\left({x}\right)\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left({x}\right) \\ $$$$\: \\ $$$$\:\:\:\:\:\int\mathrm{sec}^{\mathrm{3}} \left({x}\right){dx}\:=\:\mathrm{sec}\left({x}\right)\mathrm{tan}\left({x}\right)−\int\mathrm{sec}\left({x}\right)\mathrm{tan}^{\mathrm{2}} \left({x}\right){dx}\: \\ $$$$\:\:\:\:\:\int\mathrm{sec}^{\mathrm{3}} \left({x}\right){dx}\:=\:\mathrm{sec}\left({x}\right)\mathrm{tan}\left({x}\right)−\int\mathrm{sec}^{\mathrm{3}} \left({x}\right){dx}+\int\mathrm{sec}\left({x}\right){dx} \\ $$$$\:\:\:\:\:\mathrm{2}\int\mathrm{sec}^{\mathrm{3}} \left({x}\right){dx}\:=\:\mathrm{sec}\left({x}\right)\mathrm{tan}\left({x}\right)+\mathrm{ln}\mid\mathrm{sec}\left({x}\right)+\mathrm{tan}\left({x}\right)\mid \\ $$$$\:\:\:\:\:\int\mathrm{sec}^{\mathrm{3}} \left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\left({x}\right)\mathrm{tan}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{sec}\left({x}\right)+\mathrm{tan}\left({x}\right)\mid \\ $$$$\: \\ $$$${I}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sec}^{\mathrm{2}} \left({x}\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sec}\left({x}\right)\mathrm{tan}\left({x}\right)+\mathrm{ln}\mid\mathrm{sec}\left({x}\right)+\mathrm{tan}\left({x}\right)\mid\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} \\ $$$${I}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right]−\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{2}}×\mathrm{1}+\mathrm{ln}\mid\sqrt{\mathrm{2}}+\mathrm{1}\mid−\mathrm{1}×\mathrm{0}−\mathrm{ln}\mid\mathrm{1}+\mathrm{0}\mid\right] \\ $$$${I}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\sqrt{\mathrm{2}}+\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right] \\ $$
Answered by MJS_new last updated on 12/Dec/22
![∫((sec^2 x)/(tan x −sec x))dx=∫(dx/(sin 2x −2cos x))= [t=tan (x/2) → dx=((2dt)/(t^2 +1))] =2∫((t^2 +1)/((t−1)^3 (t+1)))dt= =(1/2)∫((1/(t−1))−(1/(t+1)))dt+∫((t+1)/((t−1)^3 ))dt= =(1/2)ln ((t−1)/(t+1)) −(t/((t−1)^2 ))= =(1/2)(ln ∣tan x −sec x∣ −(tan x + sec x)tan x)+C ⇒ answer is (1/2)ln (−1+(√2)) −((1+(√2))/2)](Q182692.png)
$$\int\frac{\mathrm{sec}^{\mathrm{2}} \:{x}}{\mathrm{tan}\:{x}\:−\mathrm{sec}\:{x}}{dx}=\int\frac{{dx}}{\mathrm{sin}\:\mathrm{2}{x}\:−\mathrm{2cos}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} \left({t}+\mathrm{1}\right)}{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}+\int\frac{{t}+\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{{t}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\mid\mathrm{tan}\:{x}\:−\mathrm{sec}\:{x}\mid\:−\left(\mathrm{tan}\:{x}\:+\:\mathrm{sec}\:{x}\right)\mathrm{tan}\:{x}\right)+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$