Question

$\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}$
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Answer 1

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Answer 2

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 :

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