5
$\begingroup$

I am not able to solve the following integration, $$\int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx.$$ I have tried in different ways but most good one I think,
\begin{align*} \int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx &= \int_{0}^{\pi} \frac{\sin x}{\sin x (1+3^{\cos x})} \\ &= \int_{-1}^{1} \frac{dz}{\sqrt{1-z^2} (1+3^{z})}.\end{align*} Now how could I proceed. Please help me.

$\endgroup$
1

2 Answers 2

15
$\begingroup$

$$I=\int_{0}^{\pi} \frac{dx}{1+3^{\cos x}}$$ Use $$\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx.$$ $$I = \int_{0}^{\pi} \frac{dx}{1+3^{-\cos x}}.$$ Adding these two we get $$2I=\int_{0}^{\pi} 1 dx\implies I=\frac{\pi}{2}$$

$\endgroup$
1
  • 1
    $\begingroup$ +1 elegant answer ........ $\endgroup$ Commented Feb 9, 2020 at 16:42
8
$\begingroup$

Just for the fun of it.

Dr Zafar Ahmed DSc provided the good solution.

For the fun of it, I did consider the more general problem of $$\int_{a}^{\pi-a} \frac{dx}{1+k^{\cos x}} $$

Expanding the integrand as a Taylor series built around $x=\frac \pi 2$ we have $$\frac{1}{1+k^{\cos x}}=\frac{1}{2}+\frac{\log (k)}{4} \left(x-\frac{\pi }{2}\right) -\frac{\log (k) \left(\log ^2(k)+2\right)}{48} \left(x-\frac{\pi }{2}\right)^3 +\frac{\log(k)\left(\log ^4(k)+5 \log ^2(k)+1\right)}{480} \left(x-\frac{\pi }{2}\right)^5 +O\left(\left(x-\frac{\pi }{2}\right)^7\right)$$ which, as expected, shows only odd powers of $\left(x-\frac{\pi }{2}\right)$.

As a result $$\int_{a}^{\pi-a} \frac{dx}{1+k^{\cos x}}= \frac{\pi }{2}-a\qquad \forall k >0$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .