prove thatint[ from (0) to(2pi) ][1 / 1+3(cos x)^2]dx = pi

ans-   baha  put    3(cos² x )=3[(1+cos2x)/2]

                                       =3/2(1+cos2x) = 3/2 + 3/2cos2x.

        THEN   cos2x = (1 - tan² x) / (1+tan² x).

        [ (1 + tan² x)dx/[5/2(1+tan² x) + 3/2(1 - tan² x)] ]

        2[ sec² x dx/ ( 5 +5tan² x + 3 - 3tan² x)

        2[ sec² x dx / ( 8+ 2tan² x)]

         [ sec² x dx/ (4 + tan² x)]            THEN let tanx = t => sec² x dx = dt

        [ dt/(4+ t² )] integrate ur ans will com