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How to evaluate the integral $$\int_{25\pi / 4}^{53\pi / 4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx\ ?$$

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  • $\begingroup$ Where did this come from? $\endgroup$ – Potato Aug 9 '13 at 21:04
  • $\begingroup$ How come you know the value? $\endgroup$ – Did Aug 10 '13 at 23:07
  • $\begingroup$ Integrals ar evaluated not solved $\endgroup$ – Arjang Aug 13 '13 at 11:16
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Let $a=\pi/4$ and $$u(x)=\frac1{(1+2^{\sin x})(1+2^{\cos x})}$$

  • If $t=2a-x$ then $(\sin(x),\cos(x))=(\cos(t),\sin(t))$ hence $$u(x)=u(t)$$
  • If $t=x-2a$ then $(\sin(x),\cos(x))=(\cos(t),-\sin(t))$ hence $$u(x)=2^{\sin(t)}u(t)$$
  • If $t=4a-x$ then $(\sin(x),\cos(x))=(\sin(t),-\cos(t))$ hence $$u(x)=2^{\cos(t)}u(t)$$
  • If $t=x-4a$ then $(\sin(x),\cos(x))=(-\sin(t),-\cos(t))$ hence $$u(x)=2^{\sin(t)}2^{\cos(t)}u(t)$$

Using each of these changes of variable $x\to t$ on the intervals $(a,2a)$, $(2a,3a)$, $(3a,4a)$ and $(4a,5a)$ respectively, one sees that the integral of $u$ on $(a,5a)$ can be written as the integral on $(0,a)$ of the sum of four functions, namely, $$ \int_a^{5a}u(x)\mathrm dx=\int_0^{a}u(t)\,(1+2^{\sin(t)}+2^{\cos(t)}+2^{\sin(t)}2^{\cos(t)})\mathrm dt $$ The integrand in $t$ simplifies into $1$, hence $$ \int_a^{5a}u(x)\mathrm dx=a $$ The same applies to the integral of $u$ on every interval $(a+4na,a+4(n+1)a)$ for some integer $n$. Since the interval $(25\pi/4,53\pi/4)$ is the union of $7$ of these (the intervals such that $6\leqslant n\leqslant12$), one gets finally $$ \int_{25\pi/4}^{53\pi/4}u(x)\mathrm dx=7a=7\pi/4 $$

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For any integrable function $f:\mathbb{R}\to\mathbb{R}$ with period $T$ and any $a,b\in\mathbb{R}$ we have $$ \int_{b}^{b+T} f(x)dx=\int_{a}^{a+T}f(x)dx\tag{1} $$ And as the consequence for $k\in\mathbb{N}$ $$ k\int_{b}^{b+T} f(x)dx=\int_{a}^{a+kT}f(x)dx\tag{2} $$ Also for any integrable $f$ we have $$ \int\limits_{-c}^c f(x)dx=\int\limits_{0}^{c}(f(x)+f(-x))dx\tag{3} $$

Now we turn to evaluation $$ \begin{align} I_1 &=\int_{25\pi/4}^{29\pi/4}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\ &=\int_{\pi/4+3\cdot 2\pi}^{5\pi/4+3\cdot 2\pi}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\ &\overset{(1)}{=}\int_{\pi/4}^{5\pi/4}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\ &=\int\limits_{\pi/4}^{3\pi/4}\left(\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}+\frac{1}{(1+2^{\cos (x+\pi/2)})(1+2^{\sin (x+\pi/2)})}\right)dx\\ &=\int\limits_{\pi/4}^{3\pi/4}\frac{dx}{1+2^{\cos x}}\\ &=\int\limits_{\pi/4}^{\pi/2}\left(\frac{1}{1+2^{\cos x}}+\frac{1}{1+2^{\cos (\pi-x)}}\right)dx\\ &=\int\limits_{\pi/4}^{\pi/2}dx\\ &=\frac{\pi}{4}\\ \end{align} $$ $$ \begin{align} I_2 &=\int_{29\pi/4}^{53\pi/4}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\ &=\int_{29\pi/4}^{29\pi/4+3\cdot 2\pi}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\ &\overset{(2)}{=}3\int_{-\pi}^{\pi}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\ &\overset{(3)}{=}3\int\limits_{0}^{\pi} \left(\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}+\frac{1}{(1+2^{\cos (-x)})(1+2^{\sin (-x)})}\right)dx\\ &=3\int\limits_{0}^{\pi}\frac{dx}{1+2^{\cos x}}\\ &=3\int\limits_{-\pi/2}^{\pi/2}\frac{dx}{1+2^{\cos (x+\pi/2)}}\\ &\overset{(3)}{=}3\int\limits_{0}^{\pi/2}\left(\frac{1}{1+2^{\cos (x+\pi/2)}}+\frac{1}{1+2^{\cos (-x+\pi/2)}}\right)dx\\ &=3\int\limits_{0}^{\pi/2}=\frac{3\pi}{2}\\ \end{align} $$ Finally, $$ I=I_1+I_2=\frac{7\pi}{4} $$

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  • $\begingroup$ I think the solution is 7*pi/4 ? $\endgroup$ – Johny Aug 10 '13 at 16:06

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