Question Number 55772 by gunawan last updated on 04/Mar/19
$$\:\underset{−\pi/\mathrm{3}} {\overset{\pi/\mathrm{3}} {\int}}\:\frac{{x}\:\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} {x}}\:{dx}\:= \\ $$
Commented by maxmathsup by imad last updated on 04/Mar/19
![let A= ∫_(−(π/3)) ^(π/3) ((xsinx)/(cos^2 x)) dx ⇒ A =2 ∫_0 ^(π/3) ((xsinx)/(cos^2 x))dx by parts u=x and v^′ =((sinx)/(cos^2 x)) A =2{ [(x/(cosx))]_0 ^(π/3) −∫_0 ^(π/3) (dx/(cosx))} =2 { (π/(3(1/2))) −∫_0 ^(π/3) (dx/(cosx))} =((4π)/3) −2 ∫_0 ^(π/3) (dx/(cosx)) but ∫_0 ^(π/3) (dx/(cosx)) =_(tan((x/2))=t ) ∫_0 ^(1/( (√3))) (1/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 )) =∫_0 ^(1/( (√3))) ((2dt)/(1−t^2 )) =∫_0 ^(1/( (√3))) { (1/(1−t)) +(1/(1+t))}dt=[ln∣((1+t)/(1−t))∣]_0 ^(1/( (√3))) =ln∣(((√3)+1)/( (√3)−1))∣ ⇒ A =((4π)/3) −2 ln((((√3)+1)/( (√3)−1))) .](https://www.tinkutara.com/question/Q55840.png)
$${let}\:\:{A}=\:\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{{xsinx}}{{cos}^{\mathrm{2}} {x}}\:{dx}\:\Rightarrow\:{A}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{{xsinx}}{{cos}^{\mathrm{2}} {x}}{dx}\:\:{by}\:{parts}\:{u}={x}\:{and}\:{v}^{'} =\frac{{sinx}}{{cos}^{\mathrm{2}} {x}} \\ $$$${A}\:=\mathrm{2}\left\{\:\:\left[\frac{{x}}{{cosx}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{dx}}{{cosx}}\right\}\:=\mathrm{2}\:\left\{\:\frac{\pi}{\mathrm{3}\frac{\mathrm{1}}{\mathrm{2}}}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{dx}}{{cosx}}\right\} \\ $$$$=\frac{\mathrm{4}\pi}{\mathrm{3}}\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{{dx}}{{cosx}}\:\:\:\:\:{but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{dx}}{{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\left\{\:\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right\}{dt}=\left[{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:={ln}\mid\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}\mid\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{4}\pi}{\mathrm{3}}\:−\mathrm{2}\:{ln}\left(\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}\right)\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19
![∫xtanxsexdx x∫tanxsecxdx−∫[(dx/dx)∫tanxsecxdx] dx xsecx−∫secxdx xsecx−ln(secx+tanx) ∣xsecx−ln(secx+tanx)∣_((−π)/3) ^(π/3) [{(π/3)sec(π/(3 ))−ln(sec(π/3)+tan(π/3))}−{((−π)/3)sec(((−π)/3))−ln(sec((−π)/3)+tan((−π)/3))}] =[{(π/3)×2−ln(2+(√3) )}−{((−2π)/3)−ln(2−(√3) )}] =((4π)/3)+ln(((2−(√3))/(2+(√3)))) pls check](https://www.tinkutara.com/question/Q55795.png)
$$\int{xtanxsexdx} \\ $$$${x}\int{tanxsecxdx}−\int\left[\frac{{dx}}{{dx}}\int{tanxsecxdx}\right]\:{dx} \\ $$$${xsecx}−\int{secxdx} \\ $$$${xsecx}−{ln}\left({secx}+{tanx}\right) \\ $$$$\mid{xsecx}−{ln}\left({secx}+{tanx}\right)\mid_{\frac{−\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} \\ $$$$\left[\left\{\frac{\pi}{\mathrm{3}}{sec}\frac{\pi}{\mathrm{3}\:}−{ln}\left({sec}\frac{\pi}{\mathrm{3}}+{tan}\frac{\pi}{\mathrm{3}}\right)\right\}−\left\{\frac{−\pi}{\mathrm{3}}{sec}\left(\frac{−\pi}{\mathrm{3}}\right)−{ln}\left({sec}\frac{−\pi}{\mathrm{3}}+{tan}\frac{−\pi}{\mathrm{3}}\right)\right\}\right] \\ $$$$=\left[\left\{\frac{\pi}{\mathrm{3}}×\mathrm{2}−{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\right\}−\left\{\frac{−\mathrm{2}\pi}{\mathrm{3}}−{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)\right\}\right] \\ $$$$=\frac{\mathrm{4}\pi}{\mathrm{3}}+{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right) \\ $$$${pls}\:{check} \\ $$
Commented by gunawan last updated on 04/Mar/19
$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{true}\:\mathrm{Sir} \\ $$$$\mathrm{Thanks} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19
$${thank}\:{you}\:{sir}\:{to}\:\:{make}\:{our}\:{brain}\:{active}\:{and}\:{agile}… \\ $$