Question Number 149805 by mnjuly1970 last updated on 07/Aug/21

Answered by Ar Brandon last updated on 07/Aug/21

I=∫_0 ^(π/2) csc^2 xln(1+2sin^2 x)dx { ((u(x)=ln(1+2sin^2 x))),((v′(x)=csc^2 x)) :}⇒ { ((u′(x)=((2sin2x)/(1+2sin^2 x)))),((v(x)=−cotx)) :} I=[−cotx∙ln(1+2sin^2 x)]_0 ^(π/2) +4∫_0 ^(π/2) ((sinxcos^2 x)/(sinx+2sin^3 x))dx =−4∫_0 ^(π/2) ((sin^2 x−1)/(2sin^2 x+1))dx=−2∫_0 ^(π/2) (1−(3/(2sin^2 x+1)))dx =−π+6∫_0 ^(π/2) ((sec^2 x)/(2tan^2 x+sec^2 x))dx=−π+6∫_0 ^(π/2) ((d(tanx))/(3tan^2 x+1)) =−π+2∙(√3)[arctan((√3)tanx)]_0 ^(π/2) =π((√3)−1)

Commented bymnjuly1970 last updated on 07/Aug/21

thx alot master...(mr brandon)

Commented byAr Brandon last updated on 07/Aug/21

You′re welcome ! How did you celebrate your birthday, Sir ? I guess july 1970 represents your month and year of birth.