IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Abhiroop

Integrate the following

cos²dx / (cos²x+4sin²x)

Plz showhow 2 solve....rates assured

vnkt.swaroop

(cos^2x)/(cos^2x+4sin^4x)

=(1)/(1+4tan^2x)

=(sec^2x)/(sec^2x+4tan^2xsec^2x)

=(sec^2x)/(1+5tan^2x+4tan^4x)

take tanx=t sec^2xdx=dt

=dt/(1+5t^2+4t^4)

now write it in the form of (a+b)^2 and try to solve it.

tell me whether i am correct or not?

reply is must.

deedee

one more method multiply and divide by -3

so u get d integral as (1/3)[ 1 +4/(3sin^2x+1) ]

nw just substitue tan(x/2)as t

bt itz just an alternate method d method by perso bove is also rite

ganapathi1810

Question On integration

ganapathi1810

(1/(1-4tan^2x)).dx
let tanx = t
sec^2x dx =dt
(1+tan^2x )dx =dt
1/((1+4t^2)*(1+t^2)).dt
now partial franctions using
4/(1+4t^2) .dt - 1/(1+t^2) .dt
let 2t= u
2dt =du
2(1/(1+u^2) .du - 1/(1+t^2) .dt
2 tan^-1u -tan^-1t
2 tan^-1(2tanx) - tan^-1(tanx)

ganapathi1810

(1/(1+4tan^2x)).dx
let tanx = t
sec^2x dx =dt
(1+tan^2x )dx =dt
1/((1+4t^2)*(1+t^2)).dt
now partial franctions using
4/(1+4t^2) .dt - 1/(1+t^2) .dt
let 2t= u
2dt =du
2(1/(1+u^2) .du - 1/(1+t^2) .dt
2 tan^-1u -tan^-1t
2 tan^-1(2tanx) - tan^-1(tanx)

determinedforiit

-1/3()

integrate individually

I1-I2

I2=

divide by

then change to from

consider tanx as t and solve.

do tell me if u got ur answer.

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