Question

Solve: $\int {{{\sin }^3}x.{{\cos }^2}xdx}$.

Answer 1

The integration that involves trigonometric functions can be solved with the help of trigonometric identities. If the exponent of the sine function in the integration is odd then we will take one sine out and will convert the rest of the sine function is the cosine function as we have done in our question.

Complete answer:
Integration is the topic of calculus. Integration is used to calculate integral and will be help of integration, we can find the value of many quantities like area, volume, displacement, and so on. Integration is of two types. The first one is definite integral and the second one is indefinite integral. In a definite integral, limits are given to us but in an indefinite integral, no limits are given to us. We can also have an integration that involves trigonometric functions and can be solved with the help of trigonometric identities. In the above question, $\sin x$ have an odd exponent so first, it will be converted into $\cos x$ and then we will do the substitution. After that, using suitable identities, the question will be solved.
This is the question of integration and we have to solve $\int {{{\sin }^3}x.{{\cos }^2}xdx}$.
Let, $I = \int {{{\sin }^3}x{{\cos }^2}xdx}$……eq(1)
We know that, according to the given identity
${\sin ^2}x + {\cos ^2}x = 1$
$\Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Now we will put the above value of ${\sin ^2}x$ in eq(1) which is as shown below.
$I = \int {{{\sin }^2}x{{\cos }^2}x\sin xdx}$
On putting the value, we get.
$I = \int {(1 - {{\cos }^2}x){{\cos }^2}x\sin xdx}$……eq(2)
This is the trigonometric integration so, in the place of $\cos x$, we have to put some another variable so that solving the question becomes easy.
Let, $\cos x = t$.
On doing the differentiation of the above expression we get the following result.
$\sin xdx = dt$
On putting this value in eq(2), the following results will be obtained.
$I = \int {(1 - {t^2}){t^2}dt}$
$\Rightarrow I = \int {({t^2} - {t^4})dt}$
Now we will do this integration separately.
$I = \int {{t^2}dt - \int {{t^4}dt} }$………eq(3) $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3}$
We know that, according to the formula of integration.
$\int {{x^n}dx = \int {\dfrac{{{x^{n + 1}}}}{{n + 1}}} } dx$
On using the above formula in eq(3).
$I = \dfrac{{{t^3}}}{3} - \dfrac{{{t^5}}}{5} + c$…….eq(4)
We have used the value of $\cos x = t$. So again putting the original value in eq(4), we get the following result.
$I = \dfrac{{{{\cos }^3}x}}{3} - \dfrac{{{{\cos }^5}x}}{5} + c$
So this is the required result of the above integration.

Note:
There is a link between differentiation and integration. Both integration and differentiation are the branches of calculus. Integration and differentiation are opposite of each other. The sum of integration of any function is equal to the sum of the individual functions of the integration.
In the given integration $\int {{{\sin }^3}x.{{\cos }^2}xdx}$, we can note that there are no limits used. If the integral does not contain any limit, then the given integral is said to be an indefinite integral. Hence $\int {{{\sin }^3}x.{{\cos }^2}xdx}$ is said to be an indefinite integral.