Question

Find ?????

Answer 1

what?

Answer 2

\[\int\limits_{ }^{}( \sqrt \cot + \sqrt tanx).dx\]

Answer 3

not a nice problem

Answer 4

ohh then plz solve

Answer 5

where did you get the question from?

Answer 6

in Sample Paper

Answer 7

that dosent tell me anything

Answer 8

can you do it again @Mimi_x3 ?

Answer 9

\[\int \sqrt{tanx}+\sqrt{cotx}dx =>\int\frac{sinx+cosx}{\sqrt{sinxcosx}} => \int\frac{sinx+cosx}{\sqrt{sin2x}}\]
\(u=sinx-cosx\) =>\(du=(sinx+cosx)dx\)
\(1-sin2x =u^2\) => \(sin2x = (1-u^2)\)
Therefore,
\[ \int\frac{1}{\sqrt{1-u^2}}du\]

Answer 10

arcsin(sinx-cosx) +const

Answer 11

i have a feeling that something is wrong :|

Answer 12

I would help if i could, sorry

Answer 13

arcsin(sinx-cosx) +const this is not answer
i write it for you to test "85
Mimi_x3" answer
but its not true

Answer 14

this integral has no answer in simple way

Answer 15

i turn it to complex form but its not easy

Answer 16

you diffrenciate this sqrt2 arcsin(sinx-cosx)
i think its correct

Answer 17

Yeah, i made a minor error; sorry!
Happens, when i dont do it on paper lol
\(sin2x = 2sinxcosx\)
\(sinxcosx = 1/2sin2x \)
\[\int\limits\frac{\sin+cosx}{\sqrt{1/2\sin2x}} =>\sqrt{2} \int\limits\frac{sinx+cosx}{\sqrt{\sin2x}} =>\sqrt{2}\int\limits\frac{1}{\sqrt{1-u^{2}}} du\]

Answer 18

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
\[ dx=2t . dt / 1+t ^{4} \]

Answer 19

NCERT example if I am recalling correctly eh ?

Answer 20

why did u re write it in that form?... it dusnt look right.......... btw, thats a good question, a challenging one

Answer 21

yeah it isz

Answer 22

@rishabh.mission: the solution i gave; is the easiest method to solve this integral.

Answer 23

ok thnku