$\int \sin mx\,dx$ 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: NCERT Class 12 Mathematics Ch 7 "Integrals", p.297_
Related questions
$\displaystyle\int_2^3 \dfrac{x\,dx}{x^2+1}$ equals$\displaystyle\int_0^{2a} f(x)\,dx$ equals $2\displaystyle\int_0^{a} f(x)\,dx$ precisely wIf $f(a+b-x)=f(x)$, then $\displaystyle\int_a^b x\,f(x)\,dx$ is equal to$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal toA rational function $\dfrac{P(x)}{Q(x)}$ is called proper when$\displaystyle\int_0^{\pi/4} \tan x\,dx$ equals$\displaystyle\int_0^{1} x e^{x^2}\,dx$ equalsThe value of $\displaystyle\int_0^{\pi/2} \log\!\left(\dfrac{4+3 in x}{4+3\cos x}\right)dx