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.