Archive for July, 2023

Integrating 1/(sin(x)+sec(x)) from 0 to pi/2

Posted: July 31, 2023 in Mathematics
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\displaystyle \int_0^{\pi/2} \frac{1}{\sin x + \sec x} \text{ d}x

Solution

\displaystyle I = \int_0^{\pi/2} \frac{\cos{x}}{\sin{x}\cos{x}+1} \text{ d}x = \int_0^{\pi/2} \frac{2\cos{x}}{2\sin{x}\cos{x}+2} \text{ d}x

\displaystyle = \int_0^{\pi/2} \frac{\cos x - \sin x}{2\sin x \cos x + 2} \text{ d}x + \int_0^{\pi/2} \frac{\cos x + \sin x}{2\sin x \cos x + 2} \text{ d}x

\displaystyle = \int_0^{\pi/2} \frac{\cos x - \sin x}{(\sin x + \cos x)^2 + 1} \text{ d}x - \int_0^{\pi/2} \frac{\cos x + \sin x}{(\sin x - \cos x)^2 - 3} \text{ d}x

Let u = sin x + cos x and v = sin x – cos x,

\displaystyle = \int_1^1 \frac{\text{d}u}{u^2+1} - \int_{-1}^1 \frac{\text{d}v}{v^2-3} = 2 \int_0^1 \frac{\text{d}v}{3-v^2}

\displaystyle = \frac{1}{\sqrt3} \int_0^1 \left(\frac{1}{\sqrt3+v} + \frac{1}{\sqrt3-v}\right) \text{d}v

\displaystyle = \frac{1}{\sqrt3}\left(\log(\sqrt3+v) - \log(\sqrt3-v)\right)\bigg|_0^1

\displaystyle = \frac{1}{\sqrt3} \left(\log(\sqrt3+1) - \log(\sqrt3-1)\right)

\displaystyle = \frac{1}{\sqrt3} \log\frac{\sqrt3+1}{\sqrt3-1}

Done.