Integrate
∫ cos(2x). ㏑(1 + tanx)
Answers
HELLO DEAR,
GIVEN:- I = ∫ cos(2x). ㏑(1 + tanx)
Integrate by parts by taking cos2x as second function & log (1 + tanx) as second function
=> log(1 + tanx). sin2x/2 - ∫1/(1 + tanx).sec²x . sin2x/2 .dx
=> log(1 + tanx).sin2x/2 - ∫tanx/(1 + tanx).dx
=> sinx.cosx.log(1 + tanx) - ∫(1+ tanx)/(1 + tanx).dx + ∫1/(1 + tanx).dx
=> sinx.cosx.log(1 + tanx) - ∫dx + ∫1/(1 + tanx).dx
=> sinx.cosx.log(1 + tanx) - x + ∫1/(1 + tanx).dx
=> sinx.cosx.log(1 + tanx) - x + ∫cosx.dx/(cosx + sinx)
=> sinx.cosx.log(1 + tanx) - x + 1/2∫(2cosx)/(cosx + sinx).dx
=> sinx.cosx.log(1 + tanx) - x + 1/2∫{(cosx + sinx) + (cosx - sinx)}/(sinx + cosx).dx
=> sinx.cosx.log(1 + tanx) - x + 1/2∫dx + 1/2∫(cosx - sinx)/(sinx + cosx).dx
[as, we know d(sinx + cosx)/dx = cosx - sinx]
=> sinx.cosx.log(1 + tanx) - 1/2x + 1/2log(sinx + cosx) + C
where, C is constant term.
I HOPE IT'S HELP YOU DEAR,
THANKS