Question

$\int_{0}^{\pi/2}\frac{dx}{1 + \sin x}$ equals

• A. 0
• B. 1
• C. –1
• D. 2

Answer 1

Answer: (B)

1

Answer 2

$I = \int_{0}^{\pi//2}\frac{dx}{\sin^{2}x ⥂ / ⥂ 2 + \cos^{2}x ⥂ / ⥂ 2 + 2\sin x ⥂ / ⥂ 2\cos x ⥂ / ⥂ 2}I = \int_{0}^{\pi/2}\frac{dx}{(\sin x ⥂ / ⥂ 2 + \cos x ⥂ / ⥂ 2)^{2}} = \int_{0}^{\pi/2}{\frac{\sec^{2}x ⥂ / ⥂ 2}{(1 + \tan x ⥂ / ⥂ 2)^{2}}dx}$Put $(1 + \tan x ⥂ / ⥂ 2) = t$ ⇒ $\frac{1}{2}\sec^{2}x ⥂ / ⥂ 2dx = dt$

∴ $I = 2{\int_{1}^{2}{\frac{dt}{t^{2}} = - 2\left\lbrack \frac{1}{t} \right\rbrack}}_{1}^{2} = - 2\left\lbrack \frac{1}{2} - \frac{1}{1} \right\rbrack = 1$