Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $

share|improve this question
9  
Correct. Thank heavens. –  Ron Gordon Feb 10 '13 at 13:15
1  
+1 for the hard math!!! –  Sigur Feb 10 '13 at 13:31
2  
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
add comment

1 Answer

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

share|improve this answer
    
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. –  gnometorule Mar 4 '13 at 3:58
    
@gnometorule As far as I can see, the question "it's correct?" and yes, indeed it is. This was an old unanswered question. I answered it to remove it from the list of unanswered questions. –  leo Mar 4 '13 at 4:03
    
To try to remove it. This answer needs at least one in order to do so –  leo Mar 4 '13 at 4:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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