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what?
\[\int\limits_{ }^{}( \sqrt \cot + \sqrt tanx).dx\]
not a nice problem

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Other answers:

ohh then plz solve
where did you get the question from?
in Sample Paper
that dosent tell me anything
can you do it again @Mimi_x3 ?
\[\int \sqrt{tanx}+\sqrt{cotx}dx =>\int\frac{sinx+cosx}{\sqrt{sinxcosx}} => \int\frac{sinx+cosx}{\sqrt{sin2x}}\] \(u=sinx-cosx\) =>\(du=(sinx+cosx)dx\) \(1-sin2x =u^2\) => \(sin2x = (1-u^2)\) Therefore, \[ \int\frac{1}{\sqrt{1-u^2}}du\]
arcsin(sinx-cosx) +const
i have a feeling that something is wrong :|
I would help if i could, sorry
arcsin(sinx-cosx) +const this is not answer i write it for you to test "85 Mimi_x3" answer but its not true
this integral has no answer in simple way
i turn it to complex form but its not easy
you diffrenciate this sqrt2 arcsin(sinx-cosx) i think its correct
Yeah, i made a minor error; sorry! Happens, when i dont do it on paper lol \(sin2x = 2sinxcosx\) \(sinxcosx = 1/2sin2x \) \[\int\limits\frac{\sin+cosx}{\sqrt{1/2\sin2x}} =>\sqrt{2} \int\limits\frac{sinx+cosx}{\sqrt{\sin2x}} =>\sqrt{2}\int\limits\frac{1}{\sqrt{1-u^{2}}} du\]
OK 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 \[ dx=2t . dt / 1+t ^{4} \]
NCERT example if I am recalling correctly eh ?
why did u re write it in that form?... it dusnt look right.......... btw, thats a good question, a challenging one
yeah it isz
@rishabh.mission: the solution i gave; is the easiest method to solve this integral.
ok thnku

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