2007-10-22 18:03:39 · 3 answers · asked by tcooper19 1 in Science & Mathematics ➔ Mathematics
I = ∫ cos ² (3x) sin (3x) dx
Let u = cos 3x
du = - 3 sin 3x dx
- du/3 = = sin 3x dx
I = (- 1/3) ∫ u ² du
I = (-1/3) (u ³ / 3) + C
I = (-1/9) (cos 3x) ³ + C
I = (-1/9) cos ³ 3x + C
2007-10-23 04:51:07 · answer #1 · answered by Como 7 · 0⤊ 0⤋
let u = cos ( 3x)
then du = - 3 sin ( 3x) dx
so multiply -3 inside the integral, , divide by - 3 outside the integral , to get
( - 1/3 ) int [ U^2 * dU ]
or ( - 1 / 3) * [ U^3] / 3 + C
( - 1/ 9 ) [ cos ( 3x) ]^3 + C
2007-10-22 18:06:36 · answer #2 · answered by Mathguy 5 · 0⤊ 0⤋
u = cos ( 3x)
du = - 3 sin ( 3x) dx
-1/3 du = sin(3x) dx
-1/3 â« u^2 du =
= -1/3 *1/3 *u^3+C
= -1/9 u^3 +C
= -1/9 *cos(3x) +C
â«cos^2 (3x) sin(3x) dx = -cos(3x) /9 +C
2007-10-22 18:11:29 · answer #3 · answered by sayamiam 6 · 0⤊ 1⤋