N Neon-Frog Member Joined Mar 27, 2003 Messages 48 Apr 30, 2003 #1 Hey We've just started the integration topic, so im not really that confident with it all. This question is really bugging me though. sin^2(x- pi/4).dx Any help appreciated Thanks guys
Hey We've just started the integration topic, so im not really that confident with it all. This question is really bugging me though. sin^2(x- pi/4).dx Any help appreciated Thanks guys
Lazarus Retired Joined Jul 6, 2002 Messages 5,965 Location CBD Gender Male HSC 2001 Apr 30, 2003 #2 Consider using the following transformation: sin<sup>2</sup> x = (1/2)(1 - cos 2x) * e.g. sin<sup>2</sup> (x - pi/4) = (1/2)(1 - cos [2x - pi/2]) * can be proved using trig rules.
Consider using the following transformation: sin<sup>2</sup> x = (1/2)(1 - cos 2x) * e.g. sin<sup>2</sup> (x - pi/4) = (1/2)(1 - cos [2x - pi/2]) * can be proved using trig rules.
N Neon-Frog Member Joined Mar 27, 2003 Messages 48 Apr 30, 2003 #3 Thanx alot I had that, either ive done something wrong in the integration process or the answer is wrong in the back. Thanks again, i'll have to go back and start from scratch to see what ive done.
Thanx alot I had that, either ive done something wrong in the integration process or the answer is wrong in the back. Thanks again, i'll have to go back and start from scratch to see what ive done.
KeypadSDM B4nn3d Joined Apr 9, 2003 Messages 2,631 Location Sydney, Inner West Gender Male HSC 2003 May 6, 2003 #4 Integral = I Sin[x-pi/4]^2dx but Sin[x]^2 = (1/2)(1 - Cos[2x]) .'. Integral = I (1/2)(1 - Cos[2x - pi/2]) = (1/4)(2x - Sin[2x-pi/2]) + c
Integral = I Sin[x-pi/4]^2dx but Sin[x]^2 = (1/2)(1 - Cos[2x]) .'. Integral = I (1/2)(1 - Cos[2x - pi/2]) = (1/4)(2x - Sin[2x-pi/2]) + c