Question Number 117944 by bemath last updated on 14/Oct/20
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{satisfies}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(\frac{\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{cos}\:\mathrm{x}}}{\:\sqrt{\mathrm{2k}}}\:\right)\:\mathrm{dx}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$
Answered by john santu last updated on 14/Oct/20
![∫ _0^(π/3) (((sin x (√(cos x)))/( (√(2k)) cos x)) ) dx = (((√2)−1)/( (√2))) ⇒(1/( (√k))) ∫_0 ^(π/3) ((sin x)/( (√(cos x)))) dx = (√2)−1 ⇒(1/( (√k) )) ∫_0 ^(π/3) ((d(cos x))/( (√(cos x)))) = 1−(√2) ⇒[ ((2 (√(cos x)))/( (√k))) ]_0 ^(π/3) = 1−(√2) ⇒(((2/( (√2)))−2)/( (√k))) = 1−(√2) ; k = ((((√2)−2)/(1−(√2))))^2 ⇒k = ((((√2)(1−(√2)))/(1−(√2))))^2 = 2](https://www.tinkutara.com/question/Q117950.png)
$$\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(\frac{\mathrm{sin}\:{x}\:\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{2}{k}}\:\mathrm{cos}\:{x}}\:\right)\:{dx}\:=\:\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{k}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{cos}\:{x}}}\:{dx}\:=\:\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{k}}\:}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\:\sqrt{\mathrm{cos}\:{x}}}\:=\:\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\left[\:\frac{\mathrm{2}\:\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{{k}}}\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} =\:\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}−\mathrm{2}}{\:\sqrt{{k}}}\:=\:\mathrm{1}−\sqrt{\mathrm{2}}\:;\:{k}\:=\:\left(\frac{\sqrt{\mathrm{2}}−\mathrm{2}}{\mathrm{1}−\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{k}\:=\:\left(\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)}{\mathrm{1}−\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$
Answered by Dwaipayan Shikari last updated on 14/Oct/20
![(1/( (√(2k))))∫_0 ^(π/3) ((sinx)/( (√(cosx))))dx=−(1/( (√(2k))))∫_1 ^(1/2) (dt/( (√t)))=(√(2/k)) [(√t)]_(1/2) ^1 (√(2/k))(1−(1/( (√2))))=1−(1/( (√2))) (2/k)=1 k=2](https://www.tinkutara.com/question/Q117951.png)
$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{{sinx}}{\:\sqrt{{cosx}}}{dx}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}}}\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dt}}{\:\sqrt{{t}}}=\sqrt{\frac{\mathrm{2}}{{k}}}\:\left[\sqrt{{t}}\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \\ $$$$\sqrt{\frac{\mathrm{2}}{{k}}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}}{{k}}=\mathrm{1} \\ $$$${k}=\mathrm{2} \\ $$
Answered by 1549442205PVT last updated on 14/Oct/20
$$\:\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(\frac{\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{cos}\:\mathrm{x}}}{\:\sqrt{\mathrm{2k}}}\:\right)\:\mathrm{dx}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Leftrightarrow\:\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(\frac{\mathrm{sinx}\:\mathrm{x}\:\sqrt{\mathrm{cos}\:\mathrm{x}}}{\:\mathrm{cosx}\sqrt{\mathrm{2k}}}\:\right)\:\mathrm{dx}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:−\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(\frac{\:\sqrt{\mathrm{cos}\:\mathrm{x}}}{\:\mathrm{cosx}\sqrt{\mathrm{2k}}}\:\right)\:\mathrm{d}\left(\mathrm{cosx}\right)\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(\frac{\:\mathrm{1}}{\:\sqrt{\mathrm{cosx}}\:\sqrt{\mathrm{2k}}}\:\right)\:\mathrm{d}\left(\mathrm{cosx}\right)\:=\:−\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{1}/\mathrm{2}} \frac{\mathrm{dt}}{\:\sqrt{\mathrm{2k}}\:\sqrt{\mathrm{t}}}\:=−\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Leftrightarrow\frac{\mathrm{2}\sqrt{\mathrm{t}}}{\:\sqrt{\mathrm{2k}}}\mid_{\mathrm{1}} ^{\mathrm{1}/\mathrm{2}} =−\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\: \\ $$$$\Leftrightarrow\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{k}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\Leftrightarrow\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{k}}}=\mathrm{1}\Leftrightarrow\mathrm{k}=\mathrm{2}\: \\ $$