Finding $\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3}).\cos (x-\frac{\pi}{6})}\mathrm{d}x$

How can I find $$\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3}).\cos (x-\frac{\pi}{6})}\mathrm{d}x$$ ? I suspect this has something simple to do with the basic definite integral properties; but can't find a way through.

• If assisted by a software, you can always integrate a rational expression of trigonometric functions. – Vim Jun 29 '15 at 6:14

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3})\cdot\cos (x-\frac{\pi}{6})}\mathrm{d}x=\int_0^{\pi/2}\frac{4}{\sqrt{3}+2\sin(2x)}dx$$

and the change of variables...

$$=\lim_{M\rightarrow\infty}\int_{\sqrt 3}^{\sqrt{3}+M}\frac{4\sqrt 3}{u(2\sqrt{3}+u)}du=\cdots$$ $$=\lim_{M\rightarrow\infty}\left[2\ln(3)+2\ln\left(\frac{M+\sqrt 3}{M+3\sqrt3}\right) \right]=2\ln(3)$$

• Which change of variables? – Jack D'Aurizio Jun 29 '15 at 12:36

$$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{dx}{\cos(x-\pi/3)\cos(x-\pi/6)}&=&\int_{0}^{\pi/2}\frac{2\,dx}{\cos(2x-\pi/2)+\cos(\pi/6)}\\&=&\int_{0}^{\pi/2}\frac{2\,dx}{\sin(2x)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{\pi/4}\frac{dx}{\sin(2x)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{\pi/4}\frac{dy}{\cos(2y)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{1}\frac{dt}{2+\left(\frac{\sqrt{3}}{2}-1\right)(1+t^2)}\\&=&8\int_{0}^{1}\frac{dt}{\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)t^2}\\&=&8\,\text{arctanh}(2-\sqrt{3})=\color{red}{2\log 3}.\end{eqnarray*}$$

Steps involved:

• $\cos(a)\cos(b)=\frac{1}{2}\left(\cos(a+b)+\cos(a-b)\right)$;
• $\sin(2x)$ is symmetric around $x=\frac{\pi}{4}$;
• we use the substitution $y=\frac{\pi}{4}-x$;
• we exploit $\cos(2y)=2\cos^2 y-1$ and substitute $y=\arctan t$;
• we finish with partial fraction decomposition.

Let $$\displaystyle y = \frac{x-\frac{\pi}{3}+x-\frac{\pi}{6}}{2} = x-\frac{\pi}{4}\Rightarrow x= \left(y+\frac{\pi}{4}\right)$$ and $dx = dy$ and changing limit

Put into $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{\cos \left(x-\frac{\pi}{3}\right)\cdot \cos \left(x-\frac{\pi}{6}\right)}dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{\cos \left(y-\frac{\pi}{12}\right)\cdot \cos \left(y+\frac{\pi}{12}\right)}dy$$

Now Using the formula $$\bullet \; \cos(A+B)\cdot \cos(A-B) = \cos^2A-\sin^2 B.$$

So Integral $$\displaystyle I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{\cos^2 y-\sin^2 \left(\frac{\pi}{12}\right)}dy = 2\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^2 y -a^2}dy = 2\int_{0}^{\frac{\pi}{4}}\frac{\sec^2 y}{1 -a^2(1+\tan^2 y)}dy$$

Where $\displaystyle a= \sin \frac{\pi}{12}$

Now Let $\tan y = t\;,$ Then $\sec^2 ydy = dt$ and Changing Limit, We get

$$\displaystyle I = 2\int_{0}^{1}\frac{1}{1-a^2(1+t^2)}dt = \frac{1}{a^2}\int_{0}^{1}\frac{1}{\left(\frac{\sqrt{1-a^2}}{a}\right)^2-t^2}dt = \frac{2}{a^2}\cdot \frac{a}{2\sqrt{1-a^2}}\ln\left|\frac{\sqrt{1-a^2}+at}{\sqrt{1-a^2}-at}\right|_{0}^{1}$$

So $$\displaystyle I = \frac{1}{a\sqrt{1-a^2}}\ln \left|\frac{\sqrt{1-a^2}+a}{\sqrt{1-a^2}-a}\right| = 2\cdot \ln \left|\frac{\cos \frac{\pi}{12}+\sin \frac{\pi}{12}}{\cos \frac{\pi}{12}-\sin \frac{\pi}{12}}\right|^2=2\ln (3)$$