If _0^/3√2k,d=1-1√2, (k0) then the value of k is:

Question

If $\displaystyle\int_0^{\pi/3}\frac{	an	heta}{\sqrt{2k\sec	heta}}\,d	heta=1-\dfrac1{\sqrt2}$, $(k>0)$ then the value of $k$ is:

Accepted Answer

$$\int\limits_0^{{\pi  \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta  = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$$

$$ \Rightarrow \,\,\int\limits_0^{\pi /3} {{{\tan \theta } \over {\sqrt {2k\sec \theta } }}} \,d\theta  = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} \right)$$

$$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos \theta } } \right)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }}$$

$$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2  + 2} \right) = 1 - {1 \over {\sqrt 2 }}$$

$$ \Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2  - 1} \right)} \over {\sqrt {2k} }} = {{\sqrt 2  - 1} \over {\sqrt 2 }}$$

$$ \Rightarrow \,\,k = 2$$

Solutions · 1