Question

Integrate the fundctions. (a)\int\frac{\sqrt{tan~x}}{sin~x~cos~x}dx​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \frac{\sqrt{tanx}}{ \: sinx \: cosx \: }dx$

Divide numerator and denominator by cos^2x, we get

$\rm \:  =  \:\displaystyle\int\sf \frac{ \dfrac{ \sqrt{tanx} }{ {cos}^{2}x } }{ \dfrac{sinx \: cosx}{ {cos}^{2}x } } \: dx$

$\rm \:  =  \:\displaystyle\int\sf \frac{ \sqrt{tanx} \: {sec}^{2} x }{tanx} \: dx$

$\rm \:  =  \:\displaystyle\int\sf \frac{\: {sec}^{2} x }{ \sqrt{tanx} } \: dx$

To solve this integral, we use method of Substitution

$\red{\rm :\longmapsto\:Put \: \sqrt{tanx} = y}$

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

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

So, above integral can now be rewritten as

$\rm \:  =  \:\displaystyle\int\sf \frac{2y \: dy}{y}$

$\rm \:  =  \:\displaystyle\int\sf 2 \: dy$

$\rm \:  =  \:2y + c$

$\rm \:  =  \:2 \sqrt{tanx} + c$

Hence,

$\rm :\longmapsto\:\boxed{ \bf{ \: \displaystyle\int\sf \frac{\sqrt{tanx}}{ \: sinx \: cosx \: }dx = 2 \sqrt{tanx} + c}}$

Additional Information :-

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