# Evaluation of definite Integral

Evaluate $$\int_{\ln(0.5)}^{\ln(2)}\left( \frac{\displaystyle\sin x \frac{\sqrt{\sin^2(\cos x)+\pi e^{(x^4)}}}{1+(xe^{\cos x}\sin x)^2}+ 2\sin(x^2+2)\arctan\left(\frac{x^3}{3}\right) } {\displaystyle 1+e^{-\frac{x^2}{2}}+x^7 \sin(-\pi x)+\frac{12}{11}|x|^{2\pi+1}} \,d x\right)$$ This is my solution, it's correct? First we observe that $\ln (0.5)=\ln \frac{1}{2}=-\ln 2$

therefore, the integral is, said $f (x)$ the integrand, $\int_{-\ln 2}^{\ln 2} f(x)\,\,dx$ that is, an integral over an interval symmetrical about the origin; without taking roads for the search of all the primitives, and by exploiting the symmetry of the interval, we check if the function is odd, in that case one can immediately conclude that the value of 'integral is $0$, then we have: therefore, the integral is, that $f (x)$ the integrand,

\begin{align} f(-x)&= \frac{\displaystyle\sin (-x)\frac{\sqrt{\sin^2(\cos (-x))+\pi e^{((-x)^4)}}}{1+((-x)e^{\cos (-x)}\sin (-x))^2}+2\sin((-x)^2+2)\arctan\left(\frac{(-x)^3}{3}\right)}{\displaystyle 1+e^{-\frac{(-x)^2}{2}}+(-x)^7 \sin(-\pi (-x))+\frac{12}{11}|(-x)|^{2\pi+1}}\\ &= \frac{\displaystyle-\sin x\frac{\sqrt{\sin^2(\cos x)+\pi e^{(x^4)}}}{1+( x e^{\cos x}\sin x )^2}-2\sin(x^2+2)\arctan\left(\frac{ x ^3}{3}\right)}{\displaystyle 1+e^{-\frac{x^2}{2}}+x^7 \sin(-\pi x )+\frac{12}{11}|x|^{2\pi+1}}\\ &= -\frac{\displaystyle \sin x\frac{\sqrt{\sin^2(\cos x)+\pi e^{(x^4)}}}{1+( x e^{\cos x}\sin x )^2}+2\sin(x^2+2)\arctan\left(\frac{ x ^3}{3}\right)}{\displaystyle 1+e^{-\frac{x^2}{2}}+x^7 \sin(-\pi x )+\frac{12}{11}|x|^{2\pi+1}}\\ &=-f(x) \end{align}

the function is therefore odd and therefore the integral is equal to $0$

-
Correct. Thank heavens. – Ron Gordon Feb 10 '13 at 13:15
+1 for the hard math!!! – Sigur Feb 10 '13 at 13:31
Some exercises in mathematics seem to have designed in the deepest and darkest dungeons of the Inquisition...but then, somehow, things turn out not that terrible. – DonAntonio Feb 10 '13 at 13:36

Yes everything is fine. ${}{}{}{}{}{}{}{}{}{}{}$