IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

magaishwarya

_{[0 ]}

^{[pi/2 ]}sin2x/[sin^4x+cos^4] dx

computer001

pls...this is not the place 4 all this
rite dinom as (sin^2x +cos^2x)^2- 2sin^2xcos^2x
then put cos2x=t
PL DO NOT RATE

studyid

hey you magaishwarya.. go to xtramarks and post this ques. there..

magaishwarya

i hav try it but the answer is not coming computer001

computer001

follow the method ive suggested..
pl no offence to u..abt wat i said previously

magaishwarya

i try in many way if you get the ans means then explain ps bro

kasirajan.1990

divide num n denom by cos^4x we get _{[ 0]}

^{[ pi/2]} { 2tanx sec^2x } / { tan^4x + 1 }

= _{[ 0]}

^{[ pi/2]}tan^(-1) (tan^2x) = pi/2

studyid

see do what computer 001 has told .. u might get 2sin2x/1+cos^2(2x)

computer001

sin2x dx/(sin^4(x) + cos^4(x))
write sin^4x +cos^4 x =(sin^2(x) +cos^2(x))^2- 2sin^2(x)cos^2(x)=1-2sin^2(x)cos^2(x)
now multiply numerator n dino by 2 so u have 2sin2x.dx/(2-(sin2x)^2) = 2sin2x/1+cos^2(2x) now put cos2x=t
u get it in the form dt/1+t^2

PL DO NOT RATE

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