If $\int_0^{\frac{\pi}{3}} \frac{ an heta}{\sqrt{2 \mathrm{ksec} heta}} \mathrm{d} heta=1-\frac{1}{\sqrt{2}},(k\gt0)$, then the value of $k$ is

Question

If $\int_0^{\frac{\pi}{3}} \frac{	an 	heta}{\sqrt{2 \mathrm{ksec} 	heta}} \mathrm{d} 	heta=1-\frac{1}{\sqrt{2}},(k\gt0)$, then the value of $k$ is

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{I}=\int_0^{\frac{\pi}{3}} \frac{	an 	heta}{\sqrt{2 \mathrm{k} \sec 	heta}} \mathrm{d} 	heta \ & =\frac{1}{\sqrt{2 \mathrm{k}}} \int_0^{\frac{\pi}{3}} \frac{\sin 	heta}{\cos 	heta} 	imes \sqrt{\cos 	heta} \mathrm{d} 	heta \ & =\frac{1}{\sqrt{2 \mathrm{k}}} \int_0^{\frac{\pi}{3}} \frac{\sin 	heta}{\sqrt{\cos 	heta}} \mathrm{~d} 	heta\end{aligned}$ $\begin{array}{ll}	herefore \quad & \begin{array}{l}	ext { Let } \cos 	heta=\mathrm{t} \ \sin 	heta \mathrm{d} 	heta=-\mathrm{dt}\end{array} \ 	herefore \quad & 	ext { when } 	heta=\frac{\pi}{3}, \mathrm{t}=\frac{1}{2} \ & 	ext { when } 	heta=0, \mathrm{t}=1 \ 	herefore \quad & I=\frac{-1}{\sqrt{2 \mathrm{k}}} \int_1^{\frac{1}{2}} \frac{1}{\sqrt{\mathrm{t}}} \mathrm{dt} \ & \quad=\frac{-1}{\sqrt{2 \mathrm{k}}} \frac{\left.\mathrm{t}^{\frac{1}{2}}ight]_1^{\frac{1}{2}}}{\frac{1}{2}}\end{array}$ $=\frac{-\sqrt{2}}{\sqrt{\mathrm{k}}}\left(\frac{1}{\sqrt{2}}-1ight)=\frac{\sqrt{2}}{\sqrt{\mathrm{k}}}\left(1-\frac{1}{\sqrt{2}}ight)$ Given that $I=\left(1-\frac{1}{\sqrt{2}}ight)$ $\Rightarrow \frac{\sqrt{2}}{\sqrt{\mathrm{k}}}=1 \Rightarrow \mathrm{k}=2$

If $\int_0^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 \mathrm{ksec} \theta}} \mathrm{d}…

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