Question Number 168225 by henderson last updated on 06/Apr/22
 ; (π/3)[ f (x) = (1/(cos x)) primitive of f(x).](Q168225.png)
$$\mathrm{hi}\:! \\ $$$$\left.{x}\:\in\:\right]\frac{\pi}{\mathrm{4}}\:;\:\frac{\pi}{\mathrm{3}}\left[\right. \\ $$$${f}\:\left({x}\right)\:=\:\frac{\mathrm{1}}{{cos}\:{x}} \\ $$$$\mathrm{primitive}\:\mathrm{of}\:{f}\left({x}\right). \\ $$
Answered by MJS_new last updated on 06/Apr/22
![∫_(π/4) ^(π/3) (dx/(cos x))= [t=tan (x/2) → dx=((2dt)/(t^2 +1))] =−2∫_(−1+(√2)) ^(1/(√3)) (dt/(t^2 −1))=∫((1/(t+1))−(1/(t+1)))dt= =[ln ∣t+1∣ −ln ∣t−1∣]_(−1+(√2)) ^(1/(√3)) = =ln (2+(√3)) −ln (1+(√2)) ∫(dx/(cos x))=ln ∣tan ((x/2)+(π/4))∣ +C= =ln ∣((cos x)/(1−sin x))∣ +C](Q168236.png)
$$\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{3}} {\int}}\:\frac{{dx}}{\mathrm{cos}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=−\mathrm{2}\underset{−\mathrm{1}+\sqrt{\mathrm{2}}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}} {\int}}\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}=\int\left(\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}= \\ $$$$=\left[\mathrm{ln}\:\mid{t}+\mathrm{1}\mid\:−\mathrm{ln}\:\mid{t}−\mathrm{1}\mid\right]_{−\mathrm{1}+\sqrt{\mathrm{2}}} ^{\mathrm{1}/\sqrt{\mathrm{3}}} = \\ $$$$=\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$$\int\frac{{dx}}{\mathrm{cos}\:{x}}=\mathrm{ln}\:\mid\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{C}= \\ $$$$=\mathrm{ln}\:\mid\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\mid\:+{C} \\ $$
Answered by floor(10²Eta[1]) last updated on 06/Apr/22
![∫_(π/4) ^(π/3) (1/(cosx))dx=∫_(π/4) ^(π/3) ((cosx)/(cos^2 x))dx=∫_(π/4) ^(π/3) ((cosx)/(1−sin^2 x))dx u=sinx⇒du=cosxdx ∫_((√2)/2) ^((√3)/2) (du/(1−u^2 ))=(1/2)∫_((√2)/2) ^((√3)/2) ((1/(1−u))+(1/(1+u)))du =−(1/2)[ln∣1−u∣]_((√2)/2) ^((√3)/2) +(1/2)[ln∣1+u∣]_((√2)/2) ^((√3)/2) −(1/2)(ln(1−((√3)/2))−ln(1−((√2)/2)))+(1/2)(ln(1+((√3)/2))−ln(1+((√2)/2))) =−(1/2)ln(((2−(√3))/2))+(1/2)ln(((2−(√2))/2))+(1/2)ln(((2+(√3))/2))−(1/2)ln(((2+(√2))/2)) =(1/2)ln(((2+(√3))/(2−(√3))))−(1/2)ln(((2+(√2))/(2−(√2)))) =ln(2+(√3))−ln(2+(√2))+ln(√2) =ln(((2+(√3))/( (√2)+1)))](Q168241.png)
$$\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \frac{\mathrm{1}}{\mathrm{cosx}}\mathrm{dx}=\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \frac{\mathrm{cosx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \frac{\mathrm{cosx}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{sinx}\Rightarrow\mathrm{du}=\mathrm{cosxdx} \\ $$$$\int_{\sqrt{\mathrm{2}}/\mathrm{2}} ^{\sqrt{\mathrm{3}}/\mathrm{2}} \frac{\mathrm{du}}{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\sqrt{\mathrm{2}}/\mathrm{2}} ^{\sqrt{\mathrm{3}}/\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{u}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\right)\mathrm{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\mid\mathrm{1}−\mathrm{u}\mid\right]_{\sqrt{\mathrm{2}}/\mathrm{2}} ^{\sqrt{\mathrm{3}}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\mid\mathrm{1}+\mathrm{u}\mid\right]_{\sqrt{\mathrm{2}}/\mathrm{2}} ^{\sqrt{\mathrm{3}}/\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{ln}\left(\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{ln}\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}−\sqrt{\mathrm{3}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)−\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)+\mathrm{ln}\sqrt{\mathrm{2}} \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by floor(10²Eta[1]) last updated on 06/Apr/22
$$\mathrm{thanks}\:\mathrm{i}\:\mathrm{corrected}\:\mathrm{it} \\ $$
Commented by MJS_new last updated on 06/Apr/22
$$\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{the}\:\mathrm{last}\:\mathrm{part}\:\mathrm{after} \\ $$$$\mathrm{inserting}\:\mathrm{the}\:\mathrm{values}.\:\mathrm{the}\:\mathrm{integral}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{ok}\:\mathrm{but}\:\mathrm{the}\:\mathrm{result}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{ln}\:\left(\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\right)\:=\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$