Question

Integrate the following function with respect to x

$\sqrt{tanx} + \sqrt{cotx}$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm ( \sqrt{tanx} + \sqrt{cotx}) \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \bigg( \sqrt{\dfrac{sinx}{cosx} } + \sqrt{\dfrac{cosx}{sinx} } \bigg) dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{sinx \: cosx} } dx$

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

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{sin2x} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - 1 + sin2x} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - (1 - sin2x)} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - ( {sin}^{2}x + {cos}^{2}x- 2 sinxcosx)} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - {(sinx - cosx)}^{2} }} dx$

Now, we use Method of Substitution, to evaluate this integral.

So, Substitute

$\purple{\rm :\longmapsto\:sinx - cosx = y}$

$\purple{\rm :\longmapsto\:cosx - ( - sinx) = \dfrac{dy}{dx} }$

$\purple{\rm :\longmapsto\:(cosx + sinx)dx = dy}$

So, above integral can be rewritten as

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

$\rm \:  =  \: \sqrt{2} {sin}^{ - 1}y + c$

$\rm \:  =  \: \sqrt{2} \: {sin}^{ - 1}(sinx - cosx) + c$

Hence,

$\boxed{\tt{ \rm \:  \displaystyle\int\rm ( \sqrt{tanx} + \sqrt{cotx})dx=\sqrt{2}{sin}^{ - 1}(sinx - cosx) + c}}$

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MORE TO KNOW

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

Answer 2

Answer:

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm ( \sqrt{tanx} + \sqrt{cotx}) \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \bigg( \sqrt{\dfrac{sinx}{cosx} } + \sqrt{\dfrac{cosx}{sinx} } \bigg) dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{sinx \: cosx} } dx$

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

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{sin2x} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - 1 + sin2x} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - (1 - sin2x)} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - ( {sin}^{2}x + {cos}^{2}x- 2 sinxcosx)} } dx$

$\rm \:  =  \: \sqrt{2} \displaystyle\int\rm \frac{sinx + cosx}{ \sqrt{1 - {(sinx - cosx)}^{2} }} dx$

Now, we use Method of Substitution, to evaluate this integral.

So, Substitute

$\purple{\rm :\longmapsto\:sinx - cosx = y}$

$\purple{\rm :\longmapsto\:cosx - ( - sinx) = \dfrac{dy}{dx} }$

$\purple{\rm :\longmapsto\:(cosx + sinx)dx = dy}$

So, above integral can be rewritten as

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

$\rm \:  =  \: \sqrt{2} {sin}^{ - 1}y + c$

$\rm \:  =  \: \sqrt{2} \: {sin}^{ - 1}(sinx - cosx) + c$

Hence,

$\boxed{\tt{ \rm \:  \displaystyle\int\rm ( \sqrt{tanx} + \sqrt{cotx})dx=\sqrt{2}{sin}^{ - 1}(sinx - cosx) + c}}$

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