How many ways to deal with the integral $\int \frac{d x}{1-\sin x \cos x}$?

Question

Multiplying both numerator and denominator of the integrand by $\sec^2 x$ yields \begin{aligned} & \int \frac{d x}{1-\sin x \cos x} \\ =& \int \frac{\sec ^{2} x}{\sec ^{2} x-\tan x} d x \\ =& \int \frac{d(\tan x)}{\tan ^{2} x-\tan x+1} \\ =& 2 \int \frac{d(2 \tan x-1)}{(2 \tan x-1)^{2}+(\sqrt{3})^{2}} \\ =& \frac{2}{\sqrt{3}} \arctan\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C . \end{aligned} Is there any simpler solution? Let me know if you have any, thank you for your attention.

Wish you enjoy the solution!

Answer 1

We could use the double-angle formula for sine,

$$\sin 2x = 2 \sin x \cos x$$

and then get

$$\newcommand{\II}{\mathcal{I}} \newcommand{\dd}{\mathrm{d}} \II := \int \frac{\dd x}{1 - \sin x \cos x} = \int \frac{\dd x}{1 - \frac 1 2 \sin 2x} =2 \int \frac{\dd x}{2 - \sin 2x}$$

Then let $u = 2x$ to get

$$\II = \int \frac{\dd u}{2 - \sin u}$$

Weierstrass substitution ($t = \tan(u/2)$) then gives

$$\II = \int \frac{1}{2 - \frac{2t}{1+t^2}} \frac{2}{1+t^2} \, \dd t = \int \frac{\dd t}{t^2 - t + 1}$$

Integral $(13)$ from this PDF then gives the answer

$$\II = \frac{2}{\sqrt 3} \arctan \left( \frac{2t - 1}{\sqrt 3} \right) + C = \frac{2}{\sqrt 3} \arctan \left( \frac{2 \tan(x) - 1}{\sqrt 3} \right) + C $$

Answer 2

another method:

$\sin(x) = \frac{tan(x)}{sec(x)}$ and $\cos(x) = \frac{1}{sec(x)}$

now you have to do u-substitution and complete the square.