3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ If assisted by a software, you can always integrate a rational expression of trigonometric functions. $\endgroup$ – Vim Jun 29 '15 at 6:14
4
$\begingroup$

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

and the change of variables...

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

$\endgroup$
  • $\begingroup$ Which change of variables? $\endgroup$ – Jack D'Aurizio Jun 29 '15 at 12:36
3
$\begingroup$

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

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.
$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.