integration

1) ∫ sin 3x cos 7x dx

= (1/2) ∫ [sin (3x + 7x) + sin (3x - 7x)] dx

= (1/2) ∫ [sin 10x + sin (- 4x)] dx

= (1/2) ∫ sin 10x - sin 4x) dx

= (1/2) [-(1/10) cos 10x + (1/4) cos 4x] + C

= (1/8) cos 4x - (1/20) cos 10x + C

2) Sub u = 2x2, then du = 4xdx and hence:

∫ x cos2 (2x2) dx = (1/4) ∫ 4x cos2 u dx

= (1/4) ∫ cos2 u du

= (1/8) ∫ (1 + cos 2u) du

= (1/8) [u + (sin 2u)/2] + C

= u/8 + (sin 2u)/16 + C

2011-01-15 11:46:59 補充：
Q2 cont'd

u/8 + (sin 2u)/16 + C

= (x^2)/4 + [sin (4x^2)]/16 + C