Qns: ∫sin 2x cos 2x dx
From math table, sin 2x = 2 sin x cos x
-> 1/2 sin 2x = sin x cos x
let x be 2x
1/2 sin 2(2x) = sin(2x) cos(2x)
-> 1/2 sin 4x = sin 2x cos 2x [same as question]
Therefore, ∫sin 2x cos 2x dx = 1/2 ∫sin 4x dx
1/2 ∫sin 4x dx = 1/2[1/4 (-cos 4x)] + c
= - 1/8 cos 4x + c
The wonder of maths.. -.-
Thursday, February 5, 2009
The reasons to forget about the unhappiness.
Sunday, February 1, 2009
The best console whenever you're down,
is when there are people who keep you alive. :D
