JAMB - Mathematics (2014 - No. 32)
Evaluate \(\int \sin 2x dx\)
cos 2x + k
\(\frac{1}{2}\)cos 2x + k
\(-\frac{1}{2}\)cos 2x + k
-cos 2x + k
Explicaţie
\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\)
\(- \frac{1}{2} \cos 2x + k\)
\(- \frac{1}{2} \cos 2x + k\)