What is the area under the curve f(x)=tanx/(secx+tanx) and the x-axis, from x=0 to x=pi

Answer 1

Area A = ∫₀^πtanx/(tanx + secx)dx = ∫₀^πsinx/(sinx + 1)dx = ∫₀^π(1 - 1/(sinx + 1)dx = ∫₀^π(1 - (1 - sinx)/cos²x)dx =

=∫₀^π(1 - sec²x + sinx/cos²x)dx = (x - tanx + secx)₀^π = (π - tanπ + secπ) - (0 - tan0 + sec0) = π - 1 - 1 = π - 2

Answer 2

The integrand is equal to sin(x) / ( 1 + sin(x) )

I used a TI-84 calculator to compute the integral with the result numerically equal to pi - 2 ~ 1.141592654