Menu Close

cotx-dx-




Question Number 94084 by seedhamaieng@gmail.com last updated on 16/May/20
∫(√(cotx))dx
$$\int\sqrt{\mathrm{cot}{x}}{dx}\: \\ $$
Commented by Kunal12588 last updated on 16/May/20
=∫(√(tan(π/2−x))) dx  =−∫(√(tan t)) dt ; with t=tan((π/2)−x)  ∫(√(tan x)) dx is very popular integral, its already  answered here ; maybe you have to dig deeper  or search in The Internet.
$$=\int\sqrt{{tan}\left(\pi/\mathrm{2}−{x}\right)}\:{dx} \\ $$$$=−\int\sqrt{{tan}\:{t}}\:{dt}\:;\:{with}\:{t}={tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\int\sqrt{{tan}\:{x}}\:{dx}\:{is}\:{very}\:{popular}\:{integral},\:{its}\:{already} \\ $$$${answered}\:{here}\:;\:{maybe}\:{you}\:{have}\:{to}\:{dig}\:{deeper} \\ $$$${or}\:{search}\:{in}\:{The}\:{Internet}. \\ $$
Commented by  M±th+et+s last updated on 16/May/20
go to Q.99340
$${go}\:{to}\:{Q}.\mathrm{99340}\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *