allery$\int \sin mx\,dx$ equals
A$-\cos mx+C$
B$\dfrac{1}{m}\cos mx+C$
C$-m\cos mx+C$
D$-\dfrac{1}{m}\cos mx+C$
Answer & Solution
Correct answer: D. $-\dfrac{1}{m}\cos mx+C$
1. Substitute $t=mx$, so $dt=m\,dx$, i.e. $dx=\dfrac{1}{m}dt$.
2. $\int \sin mx\,dx=\dfrac{1}{m}\int \sin t\,dt=-\dfrac{1}{m}\cos t$.
3. Back-substitute: $-\dfrac{1}{m}\cos mx+C$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.297_
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