Could someone post on here a semi-difficult integral if they wouldn't mind
Oh...that makes a big difference
Well lets see here $\displaystyle \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}arcos(tan(x))dx=\frac{\pi{ x}}{2}\bigg|_{\frac{-\pi}{4}}^{\frac{\pi}{4}}-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}arcsin(tan(x))dx$
Simplifying and adding the integrals we get an indentity $\displaystyle \frac{{\pi}^2}{4}=\frac{{\pi}^2}{4}$
so the answer is $\displaystyle \frac{{\pi}^2}{4}$
How did you do it? Modifying the phase shift?
Hmm...I know the way you are thinking of but I will try this
$\displaystyle \int_0^{\infty}\frac{\sin^2(x)}{x^2}dx=\frac{1}{2} \int_0^{\infty}\frac{1-\cos(2x)}{x^2}dx$
seperating we get $\displaystyle \frac{1}{2}\bigg[\int_0^{\infty}\frac{1}{x^2}dx-\int_0^{\infty}\frac{\cos(2x)}{x^2}dx\bigg]$
Now I dont feel like writing it all but I used power series and got $\displaystyle \frac{\pi}{2}$ as the answer as well
Two things...what is wrong about that...that is the answer isnt it? I even checked my calculator once I was done ...and please dont tell me I have to use the imaginary part of $\displaystyle e^{-x(a-bi)}$ I am only seventeen and a junior in highschool...dont give me this complex analysis stuff lol
Don't worry
Leibniz's integral rule
Differentiation under the integral sign - Wikipedia, the free encyclopedia
You do not need to use complex numbers
Here is an example of Leibniz's Rule in use
http://www.mathhelpforum.com/math-he...77-post19.html
And here is the complete thread: http://www.mathhelpforum.com/math-he...tegrals-2.html