Math, asked by rewaparashar5, 8 months ago

integration of √tanx / sinx.cosx

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Answered by Anonymous
20

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Answered by Anonymous
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Question : Integrate the following function

\sf\:\int\:\dfrac{\sqrt{\tan(x)}}{\sin(x)\:\cos(x)}dx

Solution :

\sf\:\int\:\dfrac{\sqrt{\tan(x)}}{\sin(x)\:\cos(x)}dx

\sf\:\implies\:\int\:\dfrac{\sqrt{\tan(x)}}{\frac({sin(x)\:\cos(x)}{\cos^2x})\:\times\:\cos^2x}dx

\sf\:\implies\:\int\:\dfrac{\sqrt{\tan(x)}}{\tan(x)}\:\times\:\sec^2(x)dx

\sf\:\implies\:\int\:\dfrac{1}{\sqrt{\tan(x)}}\:\times\:\sec^2(x)dx

If we substitute \sf\:\sqrt{\tan(x)} as t

Then we would get sec²xdx = dt

\sf\:\implies\:\int\:\dfrac{1}{t^{\frac{1}{2}}}\:\times\:dt

\sf\:\implies\:\int\:t^{\frac{-1}{2}}\:\times\:dt

\sf\:\implies\:\dfrac{t^{\frac{-1}{2}\:+\:1}}{\frac{-1}{2}\:+\:1}\:+\:c

\sf\:\implies\:\dfrac{t^{\frac{-1}{2}\:+\:1}}{\frac{1}{2}}\:+\:c

\sf\:\implies\:2t^{\frac{1}{2}}\:+\:c

\sf\:\implies\:2\sqrt{\tan(x)}\:+\:c

Trigonometric Full Table :

\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}

Trigonometric Identities :

\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ $\: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1$\end{minipage}}

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