Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 117944 by bemath last updated on 14/Oct/20

$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{satisfies}$$$$\mathrm{the}\mathrm{equation}\int {}_0^{\frac{\pi }{3}}\left(\frac{\tan \mathrm{x}\sqrt{\cos \mathrm{x}}}{\sqrt{2\mathrm{k}}}\right)\mathrm{dx}=1-\frac{1}{\sqrt{2}}$$$$$$

Answered by john santu last updated on 14/Oct/20

$$\int {}_0^{\frac{\pi }{3}}\left(\frac{\sin x\sqrt{\cos x}}{\sqrt{2k}\cos x}\right)dx=\frac{\sqrt{2}-1}{\sqrt{2}}$$$$\Rightarrow \frac{1}{\sqrt{k}}{\int }_0^{\frac{\pi }{3}}\frac{\sin x}{\sqrt{\cos x}}dx=\sqrt{2}-1$$$$\Rightarrow \frac{1}{\sqrt{k}}{\int }_0^{\frac{\pi }{3}}\frac{d\left(\cos x\right)}{\sqrt{\cos x}}=1-\sqrt{2}$$$$\Rightarrow {\left[\frac{2\sqrt{\cos x}}{\sqrt{k}}\right]}_0^{\frac{\pi }{3}}=1-\sqrt{2}$$$$\Rightarrow \frac{\frac{2}{\sqrt{2}}-2}{\sqrt{k}}=1-\sqrt{2};k={\left(\frac{\sqrt{2}-2}{1-\sqrt{2}}\right)}^2$$$$\Rightarrow k={\left(\frac{\sqrt{2}(1-\sqrt{2})}{1-\sqrt{2}}\right)}^2=2$$

Answered by Dwaipayan Shikari last updated on 14/Oct/20

$$\frac{1}{\sqrt{2k}}{\int }_0^{\frac{\pi }{3}}\frac{sinx}{\sqrt{cosx}}dx=-\frac{1}{\sqrt{2k}}{\int }_1^{\frac{1}{2}}\frac{dt}{\sqrt{t}}=\sqrt{\frac{2}{k}}{\left[\sqrt{t}\right]}_{\frac{1}{2}}^1$$$$\sqrt{\frac{2}{k}}(1-\frac{1}{\sqrt{2}})=1-\frac{1}{\sqrt{2}}$$$$\frac{2}{k}=1$$$$k=2$$

Answered by 1549442205PVT last updated on 14/Oct/20

$$\int {}_0^{\frac{\pi }{3}}\left(\frac{\tan \mathrm{x}\sqrt{\cos \mathrm{x}}}{\sqrt{2\mathrm{k}}}\right)\mathrm{dx}=1-\frac{1}{\sqrt{2}}$$$$\iff \int {}_0^{\frac{\pi }{3}}\left(\frac{\mathrm{sinx}\mathrm{x}\sqrt{\cos \mathrm{x}}}{\mathrm{cosx}\sqrt{2\mathrm{k}}}\right)\mathrm{dx}=1-\frac{1}{\sqrt{2}}$$$$-\int {}_0^{\frac{\pi }{3}}\left(\frac{\sqrt{\cos \mathrm{x}}}{\mathrm{cosx}\sqrt{2\mathrm{k}}}\right)\mathrm{d}\left(\mathrm{cosx}\right)=1-\frac{1}{\sqrt{2}}$$$$\int {}_0^{\frac{\pi }{3}}\left(\frac{1}{\sqrt{\mathrm{cosx}}\sqrt{2\mathrm{k}}}\right)\mathrm{d}\left(\mathrm{cosx}\right)=-1+\frac{1}{\sqrt{2}}$$$${\int }_1^{1/2}\frac{\mathrm{dt}}{\sqrt{2\mathrm{k}}\sqrt{\mathrm{t}}}=-1+\frac{1}{\sqrt{2}}\iff \frac{2\sqrt{\mathrm{t}}}{\sqrt{2\mathrm{k}}}{\mid }_1^{1/2}=-1+\frac{1}{\sqrt{2}}$$$$\iff \frac{\sqrt{2}}{\sqrt{\mathrm{k}}}(\frac{1}{\sqrt{2}}-1)=\frac{1}{\sqrt{2}}-1\iff \frac{\sqrt{2}}{\sqrt{\mathrm{k}}}=1\iff \mathrm{k}=2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com