Question

integrate sqrtcotx + sqrttanx dx?

Answer 1

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

Answer 2

\[=\int\limits_{}^{} \sqrt{cotx } * dx+\int\limits_{}^{} \sqrt{tanx } *dx\]

Answer 3

after 1 --
Put tan x = z 2
∴ sec2 x = 2z dz

Answer 4

\[\int\limits_{}^{}(\sqrt{\frac{ cosx }{ sinx }}+\sqrt{\frac{ sinx }{ cosx }}\]

Answer 5

\[\int\limits_{}^{}\frac{ cosx+sinx }{ \sqrt{cosx*sinx} }\]

Answer 6

Putting sinx−cosx=u,
du=(cosx+sinx)dx,
u^2=1−2sinxcosx,
sinxcosx=u^2−1^2

I=∫2√du1−u2−−−−−√=2√arcsinu+C=2√arcsin(sinx−cosx)+C

Answer 7

\[\int\limits_{}^{}\frac{ du }{ \sqrt{u^2-1} }\]
its a standard forlmula now i think

Answer 8

after the solution just replace u=sinx-cosx

Answer 9

thank you!