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

Find the value of k satisfies

the equation \int_{0}^{\frac{π}{3}} (\frac{\tan x \sqrt{\cos x}}{\sqrt{2 k}}) dx = 1 - \frac{1}{\sqrt{2}}

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

\int_{0}^{\frac{π}{3}} (\frac{\sin x \sqrt{\cos x}}{\sqrt{2 k} \cos x}) d x = \frac{\sqrt{2} - 1}{\sqrt{2}}

\Rightarrow \frac{1}{\sqrt{k}} \int_{0}^{\frac{π}{3}} \frac{\sin x}{\sqrt{\cos x}} d x = \sqrt{2} - 1

\Rightarrow \frac{1}{\sqrt{k}} \int_{0}^{\frac{π}{3}} \frac{d (\cos x)}{\sqrt{\cos x}} = 1 - \sqrt{2}

\Rightarrow {[\frac{2 \sqrt{\cos x}}{\sqrt{k}}]}_{0}^{\frac{π}{3}} = 1 - \sqrt{2}

\Rightarrow \frac{\frac{2}{\sqrt{2}} - 2}{\sqrt{k}} = 1 - \sqrt{2}; k = {(\frac{\sqrt{2} - 2}{1 - \sqrt{2}})}^{2}

\Rightarrow k = {(\frac{\sqrt{2} (1 - \sqrt{2})}{1 - \sqrt{2}})}^{2} = 2

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

\frac{1}{\sqrt{2 k}} \int_{0}^{\frac{π}{3}} \frac{s i n x}{\sqrt{c o s x}} d x = - \frac{1}{\sqrt{2 k}} \int_{1}^{\frac{1}{2}} \frac{d t}{\sqrt{t}} = \sqrt{\frac{2}{k}} {[\sqrt{t}]}_{\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{π}{3}} (\frac{\tan x \sqrt{\cos x}}{\sqrt{2 k}}) dx = 1 - \frac{1}{\sqrt{2}}

\Leftrightarrow \int_{0}^{\frac{π}{3}} (\frac{sinx x \sqrt{\cos x}}{cosx \sqrt{2 k}}) dx = 1 - \frac{1}{\sqrt{2}}

- \int_{0}^{\frac{π}{3}} (\frac{\sqrt{\cos x}}{cosx \sqrt{2 k}}) d (cosx) = 1 - \frac{1}{\sqrt{2}}

\int_{0}^{\frac{π}{3}} (\frac{1}{\sqrt{cosx} \sqrt{2 k}}) d (cosx) = - 1 + \frac{1}{\sqrt{2}}

\int_{1}^{1 / 2} \frac{dt}{\sqrt{2 k} \sqrt{t}} = - 1 + \frac{1}{\sqrt{2}} \Leftrightarrow \frac{2 \sqrt{t}}{\sqrt{2 k}} ∣_{1}^{1 / 2} = - 1 + \frac{1}{\sqrt{2}}

\Leftrightarrow \frac{\sqrt{2}}{\sqrt{k}} (\frac{1}{\sqrt{2}} - 1) = \frac{1}{\sqrt{2}} - 1 \Leftrightarrow \frac{\sqrt{2}}{\sqrt{k}} = 1 \Leftrightarrow k = 2