Question Number 32928 by Cheyboy last updated on 06/Apr/18
$$\boldsymbol{{plz}}\:\boldsymbol{{help}} \\ $$$${Evalute} \\ $$$$ \\ $$$$\underset{\pi/\mathrm{3}\:} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\sqrt{\mathrm{1}−{cosx}}}{dx} \\ $$
Commented by abdo imad last updated on 06/Apr/18
![let put I = ∫_(π/3) ^(π/4) ((sin^2 x)/(√(1−cosx))) dx I = ∫_(π/3) ^(π/4) ((4sin^2 ((x/2))cos^2 ((x/2)))/((√2) sin((x/2))))dx=(4/(√2)) ∫_(π/3) ^(π/4) sin((x/2))cos^2 ((x/2))dx =(4/(√2)) ∫_(π/4) ^(π/3) (−sin((x/2)))cos^2 ((x/2))dx =(4/(√2)) (2/3)[ cos^3 ((x/2))]_(π/4) ^(π/3) = (8/(3(√2))) (cos^3 ((π/6))−cos^3 ((π/8))) = (8/(3(√2)))( (((√3)/2))^3 −( ((√(2+(√2)))/2))^3 ) = (1/(3(√2)))( 3(√3) −((√(2+(√2))) )^3 ) .](Q32931.png)
$${let}\:{put}\:{I}\:=\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sin}^{\mathrm{2}} {x}}{\sqrt{\mathrm{1}−{cosx}}}\:{dx} \\ $$$${I}\:=\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\mathrm{4}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\sqrt{\mathrm{2}}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}=\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\left(−{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}\:\frac{\mathrm{2}}{\mathrm{3}}\left[\:{cos}^{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}}\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:=\:\frac{\mathrm{8}}{\mathrm{3}\sqrt{\mathrm{2}}}\:\left({cos}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{6}}\right)−{cos}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$=\:\frac{\mathrm{8}}{\mathrm{3}\sqrt{\mathrm{2}}}\left(\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} \:−\left(\:\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)^{\mathrm{3}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\left(\:\mathrm{3}\sqrt{\mathrm{3}}\:−\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\:\right)^{\mathrm{3}} \right)\:. \\ $$
Commented by Cheyboy last updated on 07/Apr/18
$${thankx}\:{you}\:{all}\:{for}\:{ur}\:{help} \\ $$
Answered by sma3l2996 last updated on 06/Apr/18
![I=∫_(π/3) ^(π/4) ((sin^2 x)/(√(1−cosx)))dx let t=cosx⇒dt=−sinxdx sinx=(√(1−cos^2 x))=(√(1−t^2 )) I=−∫_(1/2) ^((√2)/2) ((√(1−t^2 ))/(√(1−t)))dt=∫_((√2)/2) ^(1/2) (√(1+t))dt=(2/3)[(√((1+t)^3 ))]_((√2)/2) ^(1/2) I=(2/3)(((√(3/2)))^3 −((√((2+(√2))/2)))^3 ) etc...](Q32930.png)
$${I}=\int_{\pi/\mathrm{3}} ^{\pi/\mathrm{4}} \frac{{sin}^{\mathrm{2}} {x}}{\sqrt{\mathrm{1}−{cosx}}}{dx} \\ $$$${let}\:\:{t}={cosx}\Rightarrow{dt}=−{sinxdx} \\ $$$${sinx}=\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {x}}=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${I}=−\int_{\mathrm{1}/\mathrm{2}} ^{\sqrt{\mathrm{2}}/\mathrm{2}} \frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\sqrt{\mathrm{1}−{t}}}{dt}=\int_{\sqrt{\mathrm{2}}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \sqrt{\mathrm{1}+{t}}{dt}=\frac{\mathrm{2}}{\mathrm{3}}\left[\sqrt{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }\right]_{\sqrt{\mathrm{2}}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \\ $$$${I}=\frac{\mathrm{2}}{\mathrm{3}}\left(\left(\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\mathrm{3}} −\left(\sqrt{\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}}\right)^{\mathrm{3}} \right) \\ $$$${etc}... \\ $$
Commented by Cheyboy last updated on 07/Apr/18
$${thank}\:{you}\:{vry}\:{much}\:{sir} \\ $$
Answered by hknkrc46 last updated on 07/Apr/18
![∫_(π/3) ^(π/4) ((sin^2 x(√(1−cosx)))/(1−cosx))dx=∫_(π/3) ^(π/4) ((sinx(√(1−cosx)))/(1−cosx))sinxdx =∫_(π/3) ^(π/4) (((√(1−cos^2 x))(√(1−cosx)))/(1−cosx))sinxdx ★ 1−cosx=u⇒sinxdx=du ∧(x→(π/4)⇒u→((2−(√2))/2) ∧ x→(π/3)⇒u→(1/2)) =∫_(1/2) ^((2−(√2))/2) (((√(1−(1−u)^2 ))(√u))/u)du=∫_(1/2) ^((2−(√2))/2) ((√(2u−u^2 ))/(√u))du =∫_(1/2) ^((2−(√2))/2) (√(2−u))du=−(2/3)(2−u)^(3/2) ∣_(1/2) ^((2−(√2))/2) =−(2/3)[(2−((2−(√2))/2))^(3/2) −(2−(1/2))^(3/2) ] =−(2/3)[(((2+(√2))/2))^(3/2) −((3/2))^(3/2) ]](Q32940.png)
$$\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}\sqrt{\mathrm{1}−\mathrm{cosx}}}{\mathrm{1}−\mathrm{cosx}}\mathrm{dx}=\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sinx}\sqrt{\mathrm{1}−\mathrm{cosx}}}{\mathrm{1}−\mathrm{cosx}}\mathrm{sinxdx} \\ $$$$=\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\sqrt{\mathrm{1}−\mathrm{cosx}}}{\mathrm{1}−\mathrm{cosx}}\mathrm{sinxdx} \\ $$$$\bigstar\:\mathrm{1}−\mathrm{cosx}=\mathrm{u}\Rightarrow\mathrm{sinxdx}=\mathrm{du}\:\wedge\left(\mathrm{x}\rightarrow\frac{\pi}{\mathrm{4}}\Rightarrow\mathrm{u}\rightarrow\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\:\wedge\:\mathrm{x}\rightarrow\frac{\pi}{\mathrm{3}}\Rightarrow\mathrm{u}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}} \frac{\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{2}} }\sqrt{\mathrm{u}}}{\mathrm{u}}\mathrm{du}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}} \frac{\sqrt{\mathrm{2u}−\mathrm{u}^{\mathrm{2}} }}{\sqrt{\mathrm{u}}}\mathrm{du} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}} \sqrt{\mathrm{2}−\mathrm{u}}\mathrm{du}=−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}−\mathrm{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \mid_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left[\left(\mathrm{2}−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left[\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$
Commented by Cheyboy last updated on 07/Apr/18
$${thank}\:{u}\:{very}\:{much}\:{sir} \\ $$