$$\int\left( \sqrt{\tan x}+\sqrt{\cot x}\right)dx$$ stack exchange network stack exchange network consists of 178 q&a communities including stack overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We take, (t + 1 t) = v β (1 β 1 t2)dt = dv,&,t2 + 1 t2 = v2 β 2. Integral from 0 to pi 2 root cot x root cot x root tan x dx.
What is the integral of under root tanx? Quora
Im gonna give u the solution and i hope it is the easiest, regardless of the fact that it could be a bit lengthy.
I=β« 0 2Ο ( tanx + cotx )dx i=β« 0 2Ο ( cosxsinx + sinxcosx )dx i=β« 0 2Ο ( sinxcosx sinx+cosx )dx letz=sinxβcosx,dz=(cosx+sinx)dx z 2=sin 2x+cos 2xβ2sinxcosx z 2=1β2sinxcosx 2sinxcosx=1βz 2 sinxcosx= 21βz 2 when,x=0,t=β1;x= 2Ο ,t=1 i=β« β11.
The answer is =2sqrt(tanx)+c we need tanx=sinx/cosx sinx=cosxtanx=tanx/secx therefore, the integral is i=int(sqrt(tanx)dx)/(sinxcosx)=int(sqrt(tanx)dx)/(tanx/secx*1/secx) =int(sec^2xdx)/sqrt(tanx) let u=tanx, =>, du=sec^2xdx the integral is i=int(du)sqrt(u) =sqrt(u)/(1/2) =2sqrt(u) =2sqrt(tanx)+c = 2 β« t 2 + 1 t 4 + 1 d t. β΄ i 1 = β« 1 u2 + (β2)2 du = 1 β2 arctan( u β2). Dx = 2tdt sec2x = 2tdt 1 + tan2x = 2tdt 1 + t4.
Integral of 0 to pi/2 root (cot x) / root( cotx)+root (tan x) dx get the answers you need, now!
Integral from 0 to Ο/2 (βcot x) / [(βcot x + βtan x)] dx = login. Click hereπto get an answer to your question οΈ int^ (β(tanx)+β(cotx))dx = I hope it works out for youβ¦ thank you. Ncert solutions for class 12 physics;
I = β« ( cot x + tan x) d x.
Ncert solutions for class 12. = β« ( tan x ( 1 + cot x)) d x. (2) change into sec2x, as derivative of tan x is sec2. L e t tan x = t 2.
T β 1 t = u β (1 + 1 t2)dt = du,&,t2 + 1 t2 = u2 +2.
Ncert solutions for class 12 chemistry; T h e n, i = β« 0 Ο 4 ( sin. β d x = 2 t d t 1 + t 4. Hence, where, i 1 = β« 1 + 1 t2 t2 + 1 t2 dt.
β΄ i = β« t ( 1 + 1 t 2) Γ 2 t 1 + t 4 d t.
Sec 2 x d x = 2 t d t. Find integral of root tan x. L e t i = β« 0 Ο 4 ( tan.
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