Question

evaluate the integral tan inverse √X dx

Answer 1

Hey there!

Solution :


$Let \:I = \int tan^{-1} \sqrt{x} dx$


Let this be the equation ( a )


Put $\sqrt{x}$ = t


x = $t^{2}$


Now differentiate both sides wrt x ;


dx = 2t .dt


Substituting the value of t and dx in eq. ( a )


$I = \int tan^{-1} t . 2t .dt\\ \\ I = 2 \int t tan^{-1} .dt$


Now Integration by parts ;


$I = 2 ( tan^{-1} \int ( t .dt ) - \int ( \dfrac{d ( tan^{-1} t )}{dx} . \int ( t . dt ) ) dt\\ \\ I = 2 ( tan^{-1} t . \dfrac{t^{2}}{2} - \int ( \dfrac{1}{1 + t^{2}} . \dfrac{t^{2}}{2} ) dt\\ \\ I = 2 ( \dfrac{t^{2}}{2} tan^{-1} t - \dfrac{2}{2} \int \dfrac{t^{2}}{ 1 + t^{2}}. dt\\ \\ I = t^{2} tan^{-1} t - 1 \int \dfrac{( t^{2} + 1 ) - 1}{1 + t^{2}} .dt\\ \\ I = t^{2} tan^{-1} t - \int ( dt ) - \int ( \dfrac{1}{1 + t^{2}}. dt ) \\ \\ I = t^{2} tan^{-1} t - ( t - tan^{-1} t ) \\ \\I = t^{2} tan^{-1} t - t + tan^{-1} t + c$

Now substitute the value of t ;

$I = x tan^{-1} \sqrt{x} - \sqrt{x} + tan^{-1} \sqrt{x} + c \\ \\ I = tan^{-1} \sqrt{x} ( x + 1 ) - \sqrt{x} + c$



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