JAMB - Mathematics (2002 - No. 10)
Evaluate \(\int \sin 3x \mathrm d x\)
(2/3) cos 3x + c
(1/3) cos 3x + c
(-1/3) cos 3x + c
(-2/3) cos 3x + c
คำอธิบาย
\(\int \sin 3x \mathrm d x = - \frac{\cos 3x}{3} + c\)
= \(-\frac{1}{3} (\cos 3x) + c\)
\(\int \sin 3x \mathrm d x = - \frac{\cos 3x}{3} + c\)
= \(-\frac{1}{3} (\cos 3x) + c\)