Question

solve this

${ \int \: \sqrt{ \tan \: x } \: \: dx}$​

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm \sqrt{tanx} \: dx$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: \sqrt{tanx} = y$

$\rm \: tanx = {y}^{2}$

$\rm \: {sec}^{2}x \: dx \: = \: 2y \: dy$

$\rm \: (1 + {tan}^{2}x )\: dx \: = \: 2y \: dy$

$\rm \: (1 + {y}^{4} )\: dx \: = \: 2y \: dy$

$\rm\implies \:dx \: = \: \dfrac{2y}{ {y}^{4} + 1} \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm y \times \frac{2y}{ {y}^{4} + 1} \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{2 {y}^{2} }{ {y}^{4} + 1} \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{ {y}^{2} + {y}^{2} + 1 - 1}{ {y}^{4} + 1} \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{ ({y}^{2} + 1) + ( {y}^{2} - 1)}{ {y}^{4} + 1} \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{ {y}^{2} + 1}{ {y}^{4} + 1} \: dy \: + \: \displaystyle\int\rm \frac{ {y}^{2} - 1}{ {y}^{4} + 1} \: dy$

can be further rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2} }} \: dy \: + \: \displaystyle\int\rm \frac{1 - \dfrac{1}{ {y}^{2} }}{ {y}^{2} + \dfrac{1}{ {y}^{2} }} \: dy$

can be further rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2}} - 2 + 2} \: dy \: + \: \displaystyle\int\rm \frac{1 - \dfrac{1}{ {y}^{2} }}{ {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 - 2} \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{1}{ {y}^{2} } }{ {\bigg(y - \dfrac{1}{y} \bigg) }^{2} + 2} \: dy \: + \: \displaystyle\int\rm \frac{1 - \dfrac{1}{ {y}^{2} }}{ {\bigg(y + \dfrac{1}{y} \bigg) }^{2} - 2} \: dy$

Now, we again use method of Substitution

For first integral, Substitute

$\rm \: y - \dfrac{1}{y} = u$

$\rm \: \bigg(1 + \dfrac{1}{ {y}^{2} } \bigg)dy= du$

For Second integral, Substitute

$\rm \: y + \dfrac{1}{y} = v$

$\rm \: \bigg(1 - \dfrac{1}{ {y}^{2} } \bigg)dy= dv$

So, on substituting the values, we get

$\rm \: = \: \displaystyle\int\rm \frac{du}{ {u}^{2} + 2} \: + \: \displaystyle\int\rm \frac{dv}{ {v}^{2} - 2}$

$\rm \: = \: \displaystyle\int\rm \frac{du}{ {u}^{2} + {( \sqrt{2} )}^{2} } \: + \: \displaystyle\int\rm \frac{dv}{ {v}^{2} - {( \sqrt{2} )}^{2} }$

We know,

$\boxed{\sf{ \:\displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} \: {tan}^{ - 1} \frac{x}{a} \: + \: c \: }}$

and

$\boxed{\sf{ \:\displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{a} \: log\bigg | \frac{x - a}{x + a} \bigg| \: + \: c \: }}$

So, using these results, we get

$\rm \: = \: \dfrac{1}{ \sqrt{2} }{tan}^{ - 1} \dfrac{u}{ \sqrt{2} } + \dfrac{1}{2 \sqrt{2} }log\bigg | \dfrac{v - \sqrt{2} }{v + \sqrt{2} } \bigg| + c$

$\rm \: = \: \dfrac{1}{ \sqrt{2} }{tan}^{ - 1} \dfrac{y - \dfrac{1}{y} }{ \sqrt{2} } + \dfrac{1}{2 \sqrt{2} }log\bigg | \dfrac{y + \dfrac{1}{y} - \sqrt{2} }{y + \dfrac{1}{y} + \sqrt{2} } \bigg| + c$

$\rm \: = \: \dfrac{1}{ \sqrt{2} }{tan}^{ - 1} \dfrac{ {y}^{2} - 1}{ \sqrt{2} y} + \dfrac{1}{2 \sqrt{2} }log\bigg | \dfrac{ {y}^{2} + 1 - \sqrt{2} y}{ {y}^{2} + 1 + \sqrt{2} y} \bigg| + c$

$\rm \: = \: \dfrac{1}{ \sqrt{2} }{tan}^{ - 1} \dfrac{ tanx - 1}{ \sqrt{2} \sqrt{tanx} } + \dfrac{1}{2 \sqrt{2} }log\bigg | \dfrac{ tanx + 1 - \sqrt{2} \sqrt{tanx} }{ tanx + 1 + \sqrt{2} \sqrt{tanx} } \bigg| + c$

Hence,

$\rm \: \: \: \displaystyle\int\rm \: \sqrt{tanx} \: dx \: \\ \\ \rm \: = \: \dfrac{1}{ \sqrt{2} }{tan}^{ - 1} \dfrac{ tanx - 1}{ \sqrt{2} \sqrt{tanx} } + \dfrac{1}{2 \sqrt{2} }log\bigg | \dfrac{ tanx + 1 - \sqrt{2} \sqrt{tanx} }{ tanx + 1 + \sqrt{2} \sqrt{tanx} } \bigg| + c$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$