Question

How do you find the integral of $\int {{x^3}.\sqrt {9 - {x^2}} dx}$ ?

Answer 1

Hint : In order to determine the answer of above indefinite integral use the method of Integration by substitution by substituting $9 - {x^2}$ with ${t^2}$ then evaluate the integral by converting to some known form and finally substitute the value of t in the final answer.

Complete step-by-step answer :
Given integral $\int {{x^3}.\sqrt {9 - {x^2}} dx}$ -(1)
Here we are using Integration by substitution method to solve the above integral
Now, let’s assume $9 - {x^2} = {t^2}$ -(2)
Calculating the first derivative of the above assumed equation we get,
$\- 2xdx = 2tdt \\\ \Leftrightarrow xdx = - tdt \;$
From equation (1)
$= \int {{x^2}.\sqrt {(9 - {x^2})} x.dx}$
Now replacing $xdx = - tdt$ , ${x^2} = 9 - {t^2}$ and $9 - {x^2} = {t^2}$

$$= \int { - (9 - {t^2}){t^2}dt} \\\ = \int {({t^4} - 9{t^2})dt} \;$$

Now separating integral
$= \int {{t^4}dt} - 9\int {{t^2}dt}$
Using integration rule $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$
$= \dfrac{{{t^5}}}{5} - 9\dfrac{{{t^3}}}{3} + C$ where C is a constant
$= \dfrac{{{t^5}}}{5} - 3{t^3} + C$
Now putting back the value of $t$ back from the line no (2) in above
$= \dfrac{{{{(9 - {x^2})}^{\dfrac{5}{2}}}}}{5} - \dfrac{{{{(9 - {x^2})}^{\dfrac{3}{2}}}}}{5}+ C$ where C is the constant
Hence, Integral value of $\int {{x^3}.\sqrt {9 - {x^2}} dx}$ is equal to $\dfrac{{{{(9 - {x^2})}^{\dfrac{5}{2}}}}}{5} - \dfrac{{{{(9 - {x^2})}^{\dfrac{3}{2}}}}}{5}+ C$
So, the correct answer is “$\dfrac{{{{(9 - {x^2})}^{\dfrac{5}{2}}}}}{5} - \dfrac{{{{(9 - {x^2})}^{\dfrac{3}{2}}}}}{5}+ C$ ”.

Note : 4. Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
The symbol $\int {f(x)dx}$ is read as the indefinite integral of $f(x)$with respect to x.