IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ankit7465

cotx dx......

ankit7465

put cot x = t²

- cosec x dx = dt

(1+cot²x) dx = -dt

dx = -2t dt/1 + t⁴

now..

_{[ ]}

^{[ ]} t(-2t) dt/1+t⁴

^{am i goin right..????}

dream_iit

cosec x =/= 1 + cot^2x

cosec^2x = 1 + cot^2x

integration of cotx = log |sinx| + c

u can do this

cotx = cosx/sinx

sinx = t , cosxdx = dt

integral (1/t)

log t
= log |sinx|

fwks_phoenix

the question is root of cotx...

i suppose u have got the final ans i got, ankit.. but u'll need to proceed further n integrate.. i tried substituting again in d equation u have got.. but that's not working too...

fwks_phoenix

ok..try this way

write

tan x as (1/2){(

tan x +

cotx) +(

tan x -

cotx)}
bcuz anyway it'll give

tanx after cancelling n all...

using d formulae.. tanx=sinx/cosx and cotx=cosx/sinx
multiply numberator and denominator by

2..do this for both d operands..
and write it like

{ (sinx+cosx)^2-1}
sinx+cosx=t.. continue..

P90

express root(cotx) as root(tan(90-x)) put 90-x=t and then integrate it similar 2 root(tanx)...that will definitely be a solved example...at least in ur state board buk...

gaurav3sharma

SUBSTITUTE cot x AS t²,

THEN cosec²x dx= -2t dt,

sec²x = 1+t⁴

then the integral becomes

_{[ ]}

^{[ ]} (-2t²dt)/ 1+t⁴

which can be written as

^{[ ]} (1-t²dt)/ 1+t⁴-

^{[ ]} (t²+1dt)/ 1+t⁴

i think this becomes easier now....

as tthe above expression is very familiar type of integration.

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