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rishabh.mission
 4 years ago
Find ?????
rishabh.mission
 4 years ago
Find ?????

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

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

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

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

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

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

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

Mimi_x3
 4 years 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
 4 years ago
Best ResponseYou've already chosen the best response.0arcsin(sinxcosx) +const

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would help if i could, sorry

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

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

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

Mimi_x3
 4 years 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
 4 years 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
 4 years ago
Best ResponseYou've already chosen the best response.0NCERT example if I am recalling correctly eh ?

anonymous
 4 years 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
 4 years 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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