Integrate the following function with respect | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To integrate $\int \sin(mx) \, dx$,we use the method of substitution.Let $t = mx$. Then,differentiating both sides with respect to $x$,we get $dt

Vedclass · Accepted Answer

$-\frac{1}{m} \cos(mx) + C$

Vedclass · Answer

(A) To integrate $\int \sin(mx) \, dx$,we use the method of substitution.Let $t = mx$. Then,differentiating both sides with respect to $x$,we get $dt = m \, dx$,which implies $dx = \frac{1}{m} \, dt$.Substituting these into the integral:$\int \sin(mx) \, dx = \int \sin(t) \cdot \frac{1}{m} \, dt$$= \frac{1}{m} \int \sin(t) \, dt$$= \frac{1}{m} (-\cos(t)) + C$$= -\frac{1}{m} \cos(mx) + C$

Integrate the following function with respect to $x:$
$\sin(mx)$

Similar Questions

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} \, dx$

$\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) dx =f(x)+c$, where $c$ is the constant of integration. If $\frac{5 \pi}{2}$

$\int \frac{1}{\sin x \cdot \cos^2 x} \, dx = $

$\int {\sqrt {{{\sin }^2}x} } \,\,dx = \,;\,(x \ne n\pi ,n \in I)$

$\int {\frac{{{x^2}}}{{{x^2} + 4}}\,dx} $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Integrate the following function with respect to $x:$$\sin(mx)$