Question

Integrate the function: \[\int {\sin (mx)dx} \]

Answer 1

Hint: We solve this question by assigning the angle i.e. the term within the bracket a single variable so as to solve the integration easily. Here there is an indefinite integral used in the question, therefore we don’t have any upper limit or any lower limit to be substituted at the end in place of the variable. Also, \[m\] is a constant.
* Integration refers to the process of finding the function when its derivative is given. If we are given the derivative of a function then we can find the function by integrating both sides.
General formula for integration is \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{(n + 1)}} + C} \]
* Integration is used to find the area under the curve, volume etc.

Complete step-by-step answer:
Consider the value inside the bracket as a single variable.
Let, \[mx = t\]
Then differentiating both the sides of the equation.
\[d(mx) = d(t)\]
Since, \[m\]is a constant therefore it can be separated out from the bracket.
\[m \times dx = dt\]
Dividing both sides of the equation by \[m\]
\[
  \dfrac{{m.dx}}{m} = \dfrac{{dt}}{m} \\
  dx = \dfrac{{dt}}{m} \\
 \]
Now, we solve the integral by substituting the values \[mx = t\] and \[dx = \dfrac{{dt}}{m}\] in it.
Therefore, \[\int {\sin (mx)dx} = \int {\sin (t) \times \dfrac{{dt}}{m}} \]
Now, since the constant term is no longer with the angle along with sine, we can bring it outside of the integral using the property \[\int {k \times f(x)dx} = k\int {f(x)dx} \] where \[k\]is a constant.
Therefore, \[\int {\sin (t) \times \dfrac{{dt}}{m}} = \dfrac{1}{m}\int {\sin (t)} dt\] \[...(i)\]
Now, since we know
\[\dfrac{{d( - \cos x)}}{{dx}} = \sin x\]
Taking \[dx\] to RHS of the equation
\[d( - \cos x) = \sin xdx\]
 Therefore, taking integration on both sides of the equation gives
\[\int {d( - \cos x) = \int {\sin xdx} } \] { Since, integration and differentiation are inverse operations }
\[ - \cos x + C = \int {\sin xdx} \] Where \[C\]is constant of integration \[...(ii)\]
Substituting the value of \[ - \cos t + C = \int {\sin tdt} \] from equation \[(ii)\] in equation \[(i)\].
\[\dfrac{1}{m}\int {\sin (t)} dt = \dfrac{1}{m}( - \cos t) + C\]
                       \[ = \dfrac{{ - \cos t}}{m} + C\]
Now substitute the value \[t = mx\] back in the solution.
\[\int {\sin (mx)dx} = \dfrac{{ - \cos (mx)}}{m} + C\]

Note:
Students make the mistake of leaving the answer without converting it into its original variables, keep in mind if we ever do substitution in these kinds of questions always convert back the substitution after you get the answer because substitution is done to make calculations easier.
Also, some common integration of trigonometric terms is
1) \[\int {\sin xdx = - \cos x + C} \]
2) \[\int {\cos xdx = \sin x + C} \]