8
$\begingroup$

This question already has an answer here:

Is it possible to evaluate the following in a closed form? $$\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$


I found the above definite integral at I&S but the solution is not given. I have tried various methods but none of them lead me to the solution, I am honestly out of ideas for this one.

Any help is appreciated. Thanks!

$\endgroup$

marked as duplicate by lab bhattacharjee integration May 30 '17 at 8:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

6
$\begingroup$

We have to compute,

$$\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$ $$=$$ $$\int_{25\pi/4}^{32\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \int_{32\pi/4}^{39\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \int_{39 \pi/4}^{46\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx +\int_{46\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx = I_1+I_2+I_3+I_4 $$

Now, in $I_2$ substitute, $u=x-\frac{7 \pi}{4}$ , in $I_3$ substitute, $u = x=\frac{7\pi}{2}$ and in $I_4$ substitute, $u=x-\frac{21\pi}{4}$ .

So after substitution limits in each will integral will become same and we can directly add them. So, We have

$$\int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \frac{1}{(1+2^{\sin x+7\pi/2})(1+2^{\cos x+7\pi/2})}\,dx+ \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{(1+2^{\sin x+7\pi/4})(1+2^{\cos x+7\pi/4})}\,dx + \frac{1}{(1+2^{\sin x+21\pi/4})(1+2^{\cos x+21\pi/4})}\,dx = J_1+J_2$$

Notice that, $ \sin(a+7\pi/2) = -\cos(a) $ and $\cos(a+7\pi/2)=\sin(a) $ Using this for $J_1$ and $J_2$ we get

$$J_1 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x)}}$$ And $$ J_2 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x+7\pi/4)}} $$

Now, using well known result, that is $$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$

$$ I = J_1+J_2 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x)}} + \frac{1}{1+2^{\sin(x+7\pi/4)}} $$ Using that result, $I$ also equals $$ I = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{-\sin(x)}} + \frac{1}{1+2^{-\sin(x+7\pi/4)}} $$ $$ 2I = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} 2 dx $$ $$ I = \frac{7\pi}{4} $$


Note:: In this solution, if i have written $$ \int f(x) + \int g(x) + \cdots = I_1+I_2+\cdots $$ Then this means $$ I_1 = \int f(x) , I_2 = \int g(x) \cdots $$

$\endgroup$
  • $\begingroup$ Nice Solution.... $\endgroup$ – juantheron May 31 '15 at 7:32
  • $\begingroup$ There are hanging 'dx' terms, then there is the u=x=7*pi/2, where you meant to say u=x-7*pi/2. Could you please correct your solution? $\endgroup$ – Kugelblitz Dec 26 '16 at 12:19
3
$\begingroup$

$$I_{a,b}=\int_{a\pi}^{b\pi} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$ $$I_{25/4,53/4}=I_{0,7}=3I_{0,2}+I_{0,1}$$

$\endgroup$

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