IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - integrate 1/( (2(cosx)^2)+sinxcosx+(sinx)^2) thanks

Blaisax

integrate 1/( (2(cosx)^2)+sinxcosx+(sinx)^2) thanks

sneha.bagri

1/( (2(cosx)^2)+sinxcosx+(sinx)^2)
= 1/cos^2x +sinxcosx
=2/2cos^2x + 2sinxcosx
=2/1-cos2x+sin2x

now convert in terms of tanx

allamraju

1/1+cos²x+sinxcosx=2/1+cos2x+sin2x

converting into tanx,we get 2sec²x/1+tan²x+1-tan²x+2tanx=sec²x/1+tanx

Put tanx=t,this gives $\int$ dt/1+t=lnI1+tI=lnI1+tanxI

angel.angelina

$\int$ 1/(2(cosx^2)+sinxcosx+sinx^2= $\int$ secx^2/(2+tanx+tanx^2) LET tanx=t hence secx^2 dx=dt $\int$ 1/(2+t+t^2)= $\int$ 1/(t+2)^2+2-1/4= $\int$ 1/(t+2)^2+(sqroot3/4) applying formula for $\int$ 1/(t^2+a^2)=1/atan^-1(t/a) hence =(4/3)^1/2tan^-1((t+2)/(3/4)^1/2)) replace t to get answer. PLEASE RATE IF SATISFIED.

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1/1+cos2x+sinxcosx=2/1+cos2x+sin2x

converting into tanx,we get 2sec2x/1+tan2x+1-tan2x+2tanx=sec2x/1+tanx

Put tanx=t,this gives dt/1+t=lnI1+tI=lnI1+tanxI

1/1+cos²x+sinxcosx=2/1+cos2x+sin2x

converting into tanx,we get 2sec²x/1+tan²x+1-tan²x+2tanx=sec²x/1+tanx

Put tanx=t,this gives $\int$ dt/1+t=lnI1+tI=lnI1+tanxI