Question

integrate (√{cot x} + √{tan x})
thank you @damseldevil for answering this

can you answer this also?​

Answer 1

Question:-

$\displaystyle\sf\int\sqrt{cot\:x}+\sqrt{tan\:x}\:dx$

AnSwEr:-

$\displaystyle\sf\int\sqrt{tan\:x}(1+cot\:x)\:dx$

now use the concept of substitution

• Put tan x = t²

$\sf\implies sec^2x\:dx=2t\:dt$

$\sf\implies dx = \dfrac{2t}{1+t^4} \:\:\bf [ \because 1 + tan^2x = sec^2x]$

$\displaystyle\sf\therefore I = \int t\left(1+\dfrac{1}{t^2}\right)\dfrac{2t}{(1+t^4)}\:dt$

$\displaystyle\sf\implies I = 2 \int\dfrac{t^2+1}{t^4+1}\:dt$

on dividing numerator and denominator by t² in RHS, we get

$\displaystyle\sf I = 2\int\dfrac{\left(1+\dfrac{1}{t^2}\right)}{\left(t^2+\dfrac{1}{t^2}\right)}\:dt$

$\displaystyle\sf = 2\int \dfrac{\left(1+\dfrac{1}{t^2}\right)}{\left(t^2-\dfrac{1}{t^2}\right)^2+2}\:dt$

again use concept of substitution

• put t - 1/t = y

$\sf\implies\left(1+\dfrac{1}{t^2}\right)dt = dy$

$\displaystyle\sf \therefore I = 2\int\dfrac{dy}{y^2+(\sqrt{2})^2}$

$\sf\implies I = \sqrt{2}\:tan^{-1}\:\dfrac{y}{\sqrt{2}}+C$

$\sf = \sqrt{2}\:tan^{-1}\:\dfrac{\left(t-\dfrac{1}{t}\right)}{\sqrt{2}} + C \;\;\bf \left[ \because y = t - \dfrac{1}{t}\right]$

$\sf = \sqrt{2}\:tan^{-1}\:\left(\dfrac{t^2-1}{\sqrt{2}\:t}\right)+C$

• put value of t

$\boxed{\sf \implies I = \sqrt{2}\:tan^{-1}\left(\dfrac{tan\:x-1}{\sqrt{2}\:tan\:x}\right)+C}$

Answer 2

Answer:

THANKS ALOT DEMSEL DEVIL

WILL YOU BE MY FRIEND...