kasse Messages 383 Reaction score 1 Thread starter Nov 16, 2008 #1 If I know that [tex]\int^{\pi /2}_{0}x^2 cos^2 x dx = \frac{\pi^3}{48} - \frac{pi}{8}[/tex] can I use this to calculate [tex]\int^{\pi / 2}_{- \pi /2} u^2 cos^2 u du[/tex]?
If I know that [tex]\int^{\pi /2}_{0}x^2 cos^2 x dx = \frac{\pi^3}{48} - \frac{pi}{8}[/tex] can I use this to calculate [tex]\int^{\pi / 2}_{- \pi /2} u^2 cos^2 u du[/tex]?
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Nov 16, 2008 #2 Obviously, since the integrand is even, the latter integral is just twice the first.