Question

$\displaystyle\underline{ \bigstar \boxed{ \tt{ \int{\dfrac{ \tan \: x \sqrt{sec \: x} + sec \:x \sqrt{tan \: x} }{cos \: x}dx} }} \bigstar}$
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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{ \tan \: x \sqrt{sec \: x} + sec \:x \sqrt{tan \: x} }{cos \: x}dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \dfrac{ \dfrac{sinx}{cosx \sqrt{cosx} } + \dfrac{ \sqrt{sinx} }{cosx \sqrt{cosx} } }{cos \: x}dx \:$

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

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

$\rm \: = \: \displaystyle\int\rm \frac{sinx}{ {\bigg(cosx \bigg) }^{\dfrac{5}{2} } } \: dx + \displaystyle\int\rm \sqrt{tanx} \: {sec}^{2}x \: dx$

Now, we use Method of Substitution, we get

In first integral, We substitute

$\red{\rm :\longmapsto\:cosx = y}$

$\red{\rm :\longmapsto\: - sinx \: dx= dy}$

In second integral, we substitute

$\red{\rm :\longmapsto\: tanx = z}$

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

So, above integral can be rewritten as

$\rm \: = \: - \displaystyle\int\rm \frac{dx}{ {\bigg(x\bigg) }^{\dfrac{5}{2} } } + \displaystyle\int\rm \sqrt{z} \: dx$

$\rm \: = \: - \displaystyle\int\rm {\bigg(x \bigg) }^{ - \dfrac{5}{2} } \: dx \: + \displaystyle\int\rm {\bigg(z \bigg) }^{ \dfrac{1}{2} } \: dz$

$\rm \: = \: - \dfrac{ {\bigg(x \bigg) }^{ - \dfrac{5}{2} + 1} }{ - \dfrac{5}{2} + 1} + \dfrac{ {\bigg(z\bigg) }^{\dfrac{1}{2} + 1} }{\dfrac{1}{2} + 1} + c$

$\rm \: = \: \dfrac{ {\bigg(x \bigg) }^{ - \dfrac{3}{2}} }{\dfrac{3}{2}} + \dfrac{ {\bigg(z\bigg) }^{\dfrac{3}{2}} }{\dfrac{3}{2}} + c$

$\rm \: = \: \dfrac{2}{3} {\bigg(x \bigg) }^{ - \dfrac{3}{2}} + \dfrac{2}{3} {\bigg(z\bigg) }^{\dfrac{3}{2}} + c$

$\rm \: = \: \dfrac{2}{3} {\bigg(cosx \bigg) }^{ - \dfrac{3}{2}} + \dfrac{2}{3} {\bigg(tanx\bigg) }^{\dfrac{3}{2}} + c$

$\rm \: = \: \dfrac{2}{3cosx \sqrt{cosx} } + \dfrac{2tanx \sqrt{tanx} }{3} + c$

$\rm \: = \: \dfrac{2}{3}\bigg(secx \sqrt{secx} + tanx \sqrt{tanx} \bigg)$

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Additional Information :-

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

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$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$