Question Number 42679 by maxmathsup by imad last updated on 31/Aug/18
$${calculate}\:\:\:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\:\:\frac{{sinx}}{{cosx}\:+{tanx}}{dx}\:. \\ $$
Commented by Meritguide1234 last updated on 03/Sep/18
Commented by maxmathsup by imad last updated on 03/Sep/18
![let A = ∫_(π/4) ^(π/3) ((sinx)/(cosx +tanx)) dx ⇒ A = ∫_(π/4) ^(π/3) ((sinx cosx)/(cos^2 x +sinx))dx = ∫_(π/4) ^(π/3) ((sinx cosxdx)/(1−sin^2 x +sinx)) =_(sinx =t) ∫_(1/(√2)) ^((√3)/2) ((t dt)/(1−t^2 +t)) = −∫_(1/(√2)) ^((√3)/2) (t/(t^2 −t −1)) dt =−(1/2) ∫_(1/(√2)) ^((√3)/2) ((2t−1−1)/(t^2 −t−1))dt =−(1/2)[ ln∣t^2 −t−1∣]_(1/(√2)) ^((√3)/2) +(1/2) ∫_(1/(√2)) ^((√3)/2) (dt/(t^2 −t −1)) =−(1/2){ ln∣(3/4)−((√3)/2)−1∣−ln∣(1/2)−(1/(√2))−1∣ +(1/2) ∫_(1/(√2)) ^((√3)/2) (dt/(t^2 −t −1)) but ∫_(1/(√2)) ^((√3)/2) (dt/(t^2 −t−1))dt = ∫_(1/(√2)) ^((√3)/2) (dt/((t−(1/2))^2 −(3/4))) =_(t−(1/2)=((√3)/2) u) ∫_((2/(√3))((1/(√2))−(1/2))) ^((2/(√3))((((√3)−1)/2))) (1/((3/4)(u^2 −1)))((√3)/2)du =(4/3) ((√3)/2) ∫_((1/(√3))((√2)−1)) ^(((√3)−1)/(√3)) (du/(u^2 −1)) =(1/(√3)) ∫_(((√2)−1)/(√3)) ^(((√3)−1)/(√3)) ( (1/(u−1)) −(1/(u+1)))du =(1/(√3))[ln∣((u−1)/(u+1))∣]_(((√2)−1)/(√3)) ^(((√3)−1)/(√3)) =(1/(√3)){ ln∣ (((((√3)−1)/(√3))−1)/((((√3)−1)/(√3))+1))∣ −ln∣(((((√2)−1)/(√3))−1)/((((√2)−1)/(√3)) +1))∣} =(1/(√3)){ ln∣ (1/(2(√3)−1))∣ −ln∣(((√2)−1−(√3))/((√2)−1 +(√3)))∣ ⇒ I =−(1/2)ln((1/4)+((√3)/2))−ln((1/2)+(1/(√2))) +(1/(√3)){ ln∣(1/(2(√3)−1))∣−ln∣(((√2)−1−(√3))/((√2)−1+(√3)))∣ .](Q42844.png)
$${let}\:{A}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{sinx}}{{cosx}\:+{tanx}}\:{dx}\:\Rightarrow\:{A}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{sinx}\:{cosx}}{{cos}^{\mathrm{2}} \:{x}\:+{sinx}}{dx} \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\:\frac{{sinx}\:{cosxdx}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}\:+{sinx}}\:=_{{sinx}\:={t}} \:\:\:\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\:\:\frac{{t}\:{dt}}{\mathrm{1}−{t}^{\mathrm{2}} \:+{t}} \\ $$$$=\:−\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\:\frac{{t}}{{t}^{\mathrm{2}} −{t}\:−\mathrm{1}}\:{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\frac{\mathrm{2}{t}−\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\:{ln}\mid{t}^{\mathrm{2}} −{t}−\mathrm{1}\mid\right]_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −{t}\:−\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{ln}\mid\frac{\mathrm{3}}{\mathrm{4}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}\mid−{ln}\mid\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\mathrm{1}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −{t}\:−\mathrm{1}}\:\:{but}\right. \\ $$$$\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt}\:\:=\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{4}}}\:\:=_{{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{u}} \:\:\:\:\int_{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\right)} ^{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:\int_{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\left(\:\frac{\mathrm{1}}{{u}−\mathrm{1}}\:−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left[{ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\right]_{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left\{\:{ln}\mid\:\frac{\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}}−\mathrm{1}}{\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}}+\mathrm{1}}\mid\:−{ln}\mid\frac{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{3}}}−\mathrm{1}}{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\mathrm{1}}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left\{\:{ln}\mid\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}\mid\:−{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}−\mathrm{1}\:+\sqrt{\mathrm{3}}}\mid\:\Rightarrow\right. \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\:+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left\{\:{ln}\mid\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}\mid−{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{3}}}\mid\:.\right. \\ $$$$ \\ $$