If ∫ _0^π /3 tan⁡θ √2k sec⁡θ dθ =1- 1 √2, (k>0) , then the value of k is

$${\int }_{0}^{\pi /3}\frac{\mathrm{tan}x}{\sqrt{2k\mathrm{sec}x}}dx$$∫0π/3​2ksecx​tanx​dx

$$=\frac{1}{\sqrt{2k}} {\int }_{0}^{\pi /3}\frac{\mathrm{sin}x}{\sqrt{\mathrm{cos}x}}dx$$=2k​1​∫0π/3​cosx​sinx​dx

Put $t=\mathrm{cos}x$t=cosx

$dt= -\mathrm{sin}x dx,$dt=−sinxdx, also $x=0, t=1;x=\frac{\pi }{3}, t=\frac{1}{2}$x=0,t=1;x=3π​,t=21​

$$\therefore \frac{1}{\sqrt{2k}} \int _{1}^{1/2}\frac{-dt}{\sqrt{t}}$$∴2k​1​∫11/2​t​−dt​

$$=\frac{1}{\sqrt{2k}}{\int _{1/2}^{1}{t}^{-\frac{1}{2} }dt=\frac{1}{\sqrt{2k}}\times 2 {[{t}^{\frac{1}{2}}]}_{1/2}^{1}}_{ }$$=2k​1​∫1/21​t−21​dt=2k​1​×2[t21​]1/21​​

$$=\frac{\sqrt{2}}{\sqrt{k} }(1-\frac{1}{\sqrt{2}})=\frac{\sqrt{2}-1}{\sqrt{k}}$$=k​2​​(1−2​1​)=k​2​−1​

Given $\frac{\sqrt{2}-1}{\sqrt{k}}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$k​2​−1​=1−2​1​=2​2​−1​

$$\therefore k=2$$∴k=2