Find the value of the following integral

just put

cos x =(1-tan^2 x/2) / (1+tan^2 x/2)

u will get

(1+tan^2 x/2 ) (5+tan^2 x/2) / ((5-tan^2 x/2)

put tan x/2=t

then

1/2 [ (5+t^2) / (5-t^2) ]dt

u will easily now get the answer

edited.

ankitagg in the numerator you will get 5+tan(x/2)^2.

now you will get 5+t^2/5-t^2

put tan(x/2) =t

thus x=tan inverse of (t)

dx=2/(1+t^2)

nw d q turns out 2 b ..............

solvin it u get.................

nw rite .............

thus get ........................

so first part bcums................... u can easily integrate it :)

for d second part put t=(root 5)*siny '

thus dt=(root 5)*cosy dy

thus ur integral seond part reduces to ......................here y = dy last mein hai na woh .....

this =................

nw multiply and divide by sin y

u get ......................... here this equation is diveided by 25

nw put tany =z thus sec^2 y dy=dz

nw u get .................

nw u can write it as.........

cancellin d first term u can intergrate it easily nw

note: d answer u get multiply it by (-1) coz remeber we had taken that .......sumwer in d first part......

hope it helps

solly................................................................................................................................................................................................ 4 those bigggggggg spaces in betwn wich makes it difficult 2 read that  solution .bt that sol is perfect ........

Sir, please solve this one.not getting satisfactory answer

Hey , anyone please explain it nicely

no one in goiit to explain this question???????????????????????????????????????????

Deedee might be correct but here's a more simpler method.

Write the nr 3+2cosx=3(sin²x+cos²x)+2cosx=3sin²x+cosx(2+3cosx).Substituting this,the integrand becomes

3sin²x/(2+3cosx)²+cosx/2+3cosx

=(sinx)(1/2+3cosx)'+(1/2+3cosx)(sinx)'

=[sinx/2+3cosx]' whose integral is [sinx/(2+3cosx)]+c.

Hope you got it.