Question

What are the methods used to solve the integral $\int\limits_0^t {A\sin \left( {\omega t} \right)} {\text{ }}dt$

Answer 1

In the integral calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation.
Let $\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right)$, then $\smallint f\left( x \right)dx = F\left( x \right) + C$, where C is the integration constant.
We know, $\dfrac{{d\cos x}}{{dx}} = - \sin x$, then $\smallint \sin x{\text{ }}dx = - \cos x + C$.
A change in the variable of integration often reduces an integral to one of the fundamental integrals.
The method in which we change the variable to some other variable is called the method of substitution.
When the integrand involves some trigonometric functions, we use some well-known identities to find the integrals.

Complete step-by-step answer:
Step 1: Substitute
$I = \int\limits_0^t {A\sin \left( {\omega t} \right)} {\text{ }}dt$
Take $\omega t = x$
Differentiating both sides.
$\omega \;dt = dx \\\ \Rightarrow dt = \dfrac{{dx}}{\omega } \\\$
Step 2: Change of limits
$t = 0 \to x = 0 \\\ t = t \to x = \omega t \\\$
Hence, the upper limit of the integral is $\omega t$, and lower limit is $0$.
$\therefore I = \dfrac{1}{\omega }\int\limits_0^\omega {A\sin x} {\text{ }}dx$
$\Rightarrow \dfrac{1}{\omega } \times A\left. {\left[ { - \cos x} \right]} \right|_0^{\omega t}$
Step 3: solve the limits
$I = \dfrac{1}{\omega } \times A\left[ { - \cos \left( {\omega t} \right) + \cos 0} \right] \\\ \Rightarrow - \dfrac{A}{\omega }\cos \left( {\omega t} \right) \\\$
Final answer: The substitution method is mainly used to solve the integral $\int\limits_0^t {A\sin \left( {\omega t} \right)} {\text{ }}dt$.

Note: After substituting the variables, you can also avoid changing the limits. Solve the integral without taking limits into considerate, before the final answer, substitute it (variable) back and solve with previous limits.
Steps are shown below:
$I = \dfrac{1}{\omega }\int {A\sin x{\text{ }}dx}$
$\Rightarrow \dfrac{1}{\omega }A\left( { - \cos x} \right)$
We know, $x = \omega t$
$\because I = \dfrac{1}{\omega }A\left( { - \cos \omega t} \right)_0^t$
For variable $t$, limits were $0 \to t$.
Thus, solving limits we get:
$I = \dfrac{1}{\omega } \times A\left[ { - \cos \left( {\omega t} \right) + \cos 0} \right] \\\ \Rightarrow - \dfrac{A}{\omega }\cos \left( {\omega t} \right) \\\$
Students must understand that the variable $t$in the function $A\sin \left( {\omega t} \right)$ and the variable $t$in the limits $\int\limits_0^t {A\sin \left( {\omega t} \right)} {\text{ }}dt$are both different even though they are same variable.
Variable $t$in the function means that the function is dependent on $t$, we differentiate or integrate the function by $dt$.
Variable $t$in the limits is only signifying the upper limits or limits of the function.