## rishabh.mission 3 years ago Find ?????

1. anonymous

what?

2. rishabh.mission

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

3. UnkleRhaukus

not a nice problem

4. rishabh.mission

ohh then plz solve

5. UnkleRhaukus

where did you get the question from?

6. rishabh.mission

in Sample Paper

7. UnkleRhaukus

that dosent tell me anything

8. UnkleRhaukus

can you do it again @Mimi_x3 ?

9. 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, $\int\frac{1}{\sqrt{1-u^2}}du$

10. amoodarya

arcsin(sinx-cosx) +const

11. Mimi_x3

i have a feeling that something is wrong :|

12. anonymous

I would help if i could, sorry

13. amoodarya

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

14. amoodarya

this integral has no answer in simple way

15. amoodarya

i turn it to complex form but its not easy

16. amoodarya

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

17. Mimi_x3

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

18. rishabh.mission

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 $dx=2t . dt / 1+t ^{4}$

19. shubhamsrg

NCERT example if I am recalling correctly eh ?

20. anonymous

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

21. rishabh.mission

yeah it isz

22. Mimi_x3

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

23. rishabh.mission

ok thnku