Question

Prove that $\int\limits_0^t {A\sin wtdt} = A\left[ {\dfrac{{ - \cos (wt)}}{w}} \right] _0^t = \dfrac{A}{w}\left( {\cos (wt) - 1} \right)$ .

Answer 1

Hint : First we know an integration is the inverse process of differentiation. Then understand the standard rules of integration. After that integrate the given integrand with respect to the independent variable keeping the other variables as a constant.

Complete step-by-step answer :
Integration rules: Suppose $f$ and $g$ are two integral functions and integrable with respect to $x$ and $k$ is any constant non zero value then following rules are holds
Multiplication by constant: integration of constant time the function is equal to constant time the integration of the function.
i.e., $\int {\left( {kf} \right)} dx = k.\int {fdx} + C$
Sum rule: Integration of sum of two functions is equal to the sum of the integrations of the two functions.
i.e., $\int {\left( {f + g} \right)} dx = \int {fdx} + \int {gdx} + C$
Difference rule: Integration of difference of two functions is equal to the difference of the integrations of the two functions.
i.e., $\int {\left( {f - g} \right)} dx = \int {fdx} - \int {gdx} + C$ .
Given $\int\limits_0^t {A\sin wtdt}$ ---(1)
Let $I = \int {A\sin \left( {wt} \right)dt}$
Using the integrating rule, we get
$I = A\int {\sin \left( {wt} \right)dt}$
$\int {A\sin \left( {wt} \right)dt = } A\left[ {\dfrac{{ - \cos (wt)}}{w}} \right] + C$
Where $C$ is an integration constant.
Since here $t$ is a time, we know that time always starts from zero. Hence $t$ varies from $0$ to $t$
Then we consider $\int\limits_0^t {A\sin wtdt}$
Using the integration rule, we get
$\int {A\sin \left( {wt} \right)dt = } A\left[ {\dfrac{{ - \cos (wt)}}{w}} \right] _0^t$
Applying the upper and lower limit, we get
$\int\limits_0^t {A\sin wtdt} = A\left[ {\dfrac{{ - \cos (wt)}}{w} - \left( {\dfrac{{ - \cos (0)}}{w}} \right)} \right]$
Simplify the RHS of the above equation we get
$\int\limits_0^t {A\sin wtdt} = \dfrac{A}{w}\left( {\cos (wt) - 1} \right)$ .

Note : Note that there are some methods to evaluate the given integrand. Substitution method and integration by parts are two important methods. The concept of integration has developed to solve the following types of problems. To find the problem function, when its derivatives are given, and to find the area bounded by the graph of a function under certain constraints.