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rishabh.mission
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{ }^{}( \sqrt \cot + \sqrt tanx).dx\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0not a nice problem

rishabh.mission
 one year ago
Best ResponseYou've already chosen the best response.0ohh then plz solve

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0where did you get the question from?

rishabh.mission
 one year ago
Best ResponseYou've already chosen the best response.0in Sample Paper

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0that dosent tell me anything

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0can you do it again @Mimi_x3 ?

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.5\[\int \sqrt{tanx}+\sqrt{cotx}dx =>\int\frac{sinx+cosx}{\sqrt{sinxcosx}} => \int\frac{sinx+cosx}{\sqrt{sin2x}}\] \(u=sinxcosx\) =>\(du=(sinx+cosx)dx\) \(1sin2x =u^2\) => \(sin2x = (1u^2)\) Therefore, \[ \int\frac{1}{\sqrt{1u^2}}du\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0arcsin(sinxcosx) +const

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.5i have a feeling that something is wrong :

youarestupid
 one year ago
Best ResponseYou've already chosen the best response.0I would help if i could, sorry

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0arcsin(sinxcosx) +const this is not answer i write it for you to test "85 Mimi_x3" answer but its not true

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0this integral has no answer in simple way

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0i turn it to complex form but its not easy

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0you diffrenciate this sqrt2 arcsin(sinxcosx) i think its correct

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.5Yeah, i made a minor error; sorry! Happens, when i dont do it on paper lol \(sin2x = 2sinxcosx\) \(sinxcosx = 1/2sin2x \) \[\int\limits\frac{\sin+cosx}{\sqrt{1/2\sin2x}} =>\sqrt{2} \int\limits\frac{sinx+cosx}{\sqrt{\sin2x}} =>\sqrt{2}\int\limits\frac{1}{\sqrt{1u^{2}}} du\]

rishabh.mission
 one year ago
Best ResponseYou've already chosen the best response.0OK but can i solve this in a different method like can i Re write as \[\int\limits_{}^{} \sqrt{ \tan x} (1+\cot x )dx\] and then Put tan x = t², so that sec²x dx = 2t dt Or \[ dx=2t . dt / 1+t ^{4} \]

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0NCERT example if I am recalling correctly eh ?

Tushara
 one year ago
Best ResponseYou've already chosen the best response.0why did u re write it in that form?... it dusnt look right.......... btw, thats a good question, a challenging one

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.5@rishabh.mission: the solution i gave; is the easiest method to solve this integral.
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