Re: MX2 Integration Marathon
Looks like a hard integral. I did IBP and got \frac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} cosec^{2}x dx which doesn't really help.
Re: MX2 Integration Marathon
Not sure if this is correct because it looks really ugly...
$Let$ I=\int \frac{sin^{-1}(e^{x})}{e^{x}}
Let \frac{1}{e^{x}}=u\therefore dx=-due^{x}
=-\int sin^{-1}(\frac{1}{u})du
Let \frac{1}{u}=v\therefore du=-\frac{dv}{v^{2}}
\therefore I=\int...
Re: HSC 2013 4U Marathon
Good job!
By the way, \int_{0}^{\infty }sin(x^{2})dx=\frac{1}{\sqrt{\pi }}\int_{0}^{\infty }\frac{1}{x^{4}+1}dx=\frac{\sqrt{\pi} }{2\sqrt{2}}
No problem, I will post there next time.
Re: HSC 2013 4U Marathon
The final result is nice though. Allows you to evaluate a sine Fresnel Integral without using complex analysis (you need Gamma function though).
Re: HSC 2013 4U Marathon
You can't use that substitution because you can't get rid off the x after you sub dx back.
The only substitution you can use is x=tanx; \int_{0}^{\infty }\frac{sec^{2}\theta }{tan^{4}\theta+1}d\theta which doesn't really help.