Showing posts with label Integral of $sin^2x$. Show all posts
Showing posts with label Integral of $sin^2x$. Show all posts

Monday, 30 May 2016

Monday, 9 May 2016

Grove HSC 3U Ex 5.11 Q1c

Grove HSC 5.11 q2c


You got the sign wrong, its $\Large x-{\sin(2x)\over 2}$ on line 3.
Line 4 correction: $\ldots \Large \left[ { \pi^2\over 2}-{\pi^2\over 8} \right]$