`int_(0)^((pi)/(2))(cos^(2)xdx)/(cos^(2)x + 4sin^(2)x)`

Explanation

माना `I=int_(0)^(pi//2)(cos^(2)x)/(cos^(2)x+4sin^(2)x)dx" ...(1)"`
`=int_(0)^(pi//2)(cos^(2)x)/(4-3cis^(2)x)dx`
`" "(because sin^(2)x=1-cos^(2)x)`
`=-(1)/(3)int_(0)^(pi//2)(4-3cos^(2)x-4)/(4-3cos^(2)x)dx`
`=-(1)/(3)int_(0)^(pi//2)(1-(4)/(4-3cos^(2)x))dx`
`=-(1)/(3)[x]_(0)^(pi//2)+(4)/(3)(Sec^(2)x)/(4sec^(2)x-3)dx`
(दूसरे समाकलन में अंश और हर में `cos^(2)x` से भाग देने पर )
`{:("माना ", " "tan x= t ),(rArr.," "sec^(2)xdx=dt),("तथा ",x=0 " "rArr" " t=0),("और ",x=((pi)/(2))" "rArr" "t=oo):}`
`therefore" "I=-(pi)/(6)+(4)/(3)int_(0)^(oo)(dt)/(4(1+t^(2))-3)`
`=-(pi)/(6)+(4)/(3)int_(0)^(oo)(dt)/(4t^(2)+1)`
`=-(pi)/(6)+(4)/(3)xx(1)/(4)int_(0)^(oo)(dt)/(t^(2)+((1)/(2))^(2))`
`=-(pi)/(6)+(1)/(3).2[tan^(-1).(t)/((1)/(2))]_(0)^(oo)`
`=-(pi)/(6)+(2)/(3)[tan^(-2)2t]_(0)^(oo)`
`=-(pi)/(6)+(2)/(3)(tan^(-1)oo-tan^(-1)0)`
`=-(pi)/(6)+(2)/(3)((pi)/(2)-0)=-(pi)/(6)+(pi)/(3)=(pi)/(6)`