Question

Evaluate the indefinite integral $\int {\sqrt {10x - {x^2}} dx} $ .

Answer 1

Hint: We need to integrate the given integral, for that first we’ll complete the square, and then using some substitution we will integrate this integral. There are two types of integral; definite and indefinite integral. Here the given is indefinite integral since there is no limit. Here we are going to apply the substitution method and we need to transform the given integral into a simpler form of integral by substituting the independent variable.

Formula to be used:
$\int {adx = ax + c} $
$\int {\cos pxdx = \dfrac{{\sin px}}{p} + c} $
$\sin \left( {\arcsin (x)} \right) = x$
Half angle formulae: $\cos 2x = 2\cos x - 1$
Identity: ${\sin ^2}x + {\cos ^2}x = 1$
${(a - b)^2} = {a^2} - 2ab + {b^2}$

Complete answer:
Given integral: $\int {\sqrt {10x - {x^2}} dx} $
To integrate the given integral, we need to complete the square.
For that, first take negative sign common so that coefficient of ${x^2}$ becomes positive.
We get, $\int {\sqrt { - \left( {{x^2} - 10x} \right)} dx} $ .
Now, Adding and subtracting square of $\dfrac{1}{2} \times 10 = 5$ simultaneously,
we get, $\int {\sqrt { - \left( {{x^2} - 10x + {5^2} - {5^2}} \right)} dx} $
$ = \int {\sqrt { - \left( {{x^2} - 2\left( 5 \right)x + {5^2} - {5^2}} \right)} dx} $ .
$ = \int {\sqrt { - \left( {{{\left( {x - 5} \right)}^2} - {5^2}} \right)} dx} $ (Here we are using the identity ${(a - b)^2} = {a^2} - 2ab + {b^2}$)
$ = \int {\sqrt {25 - {{\left( {x - 5} \right)}^2}} dx} $
Now, substituting $x - 5 = u$ , then, on differentiating both sides, we have, $dx = du$ . The equation becomes $ = \int {\sqrt {25 - {u^2}} du} $
Now, substitute $u = 5\sin v$ , then on differentiating both sides, we have, $du = 5\cos vdv$ .
The equation becomes
 $\int {\sqrt {25 - {{\left( {5\sin v} \right)}^2}} \left( {5\cos v} \right)dv} $
$ = \int {\sqrt {25 - 25{{\sin }^2}v} \left( {5\cos v} \right)dv} $ .
Taking $25$ outside the root, we get
$ = \int {5\sqrt {1 - {{\sin }^2}v} \left( {5\cos v} \right)dv} $
$ = \int {\sqrt {1 - {{\sin }^2}v} \left( {25\cos v} \right)dv} $
Now, using the identity, ${\sin ^2}x + {\cos ^2}x = 1$ , the equation becomes,
 $ = \int {\sqrt {{{\cos }^2}v} \left( {25\cos v} \right)dv} $ .
On simplifying, we get,
$ = \int {25{{\cos }^2}vdv} $ .
Now, using the half-angle formula ${\cos ^2}v$ can be written as $\dfrac{{1 + \cos 2v}}{2}$
$ = 25\int {\left( {\dfrac{{1 + \cos 2v}}{2}} \right)dv} $
On integrating, we get, $\dfrac{{25}}{2}\left( {v + \dfrac{{\sin 2v}}{2}} \right) + c$ , where, $c$ is the constant of integration.
Finally, substitute back the values of $v = \arcsin \dfrac{u}{5}$ , we get,
\[ = \dfrac{{25}}{2}\left( {\left( {arc\sin \left( {\dfrac{u}{5}} \right) + \dfrac{{\sin \left( {2\arcsin \left( {\dfrac{u}{5}} \right)} \right)}}{2}} \right)} \right) + c\]
$ = \dfrac{{25}}{2}\left( {\arcsin \left( {\dfrac{u}{5}} \right) + \dfrac{u}{5}} \right) + c$
Now, substitute the values of $u = x - 5$ , we get,
$ = \dfrac{{25}}{2}\left( {\arcsin \left( {\dfrac{{x - 5}}{5}} \right) + \dfrac{{x - 5}}{5}} \right) + c$ .
On simplifying, we get,
$ = \dfrac{{25}}{2}\arcsin \left( {\dfrac{{x - 5}}{5}} \right) + \dfrac{5}{2}\left( {x - 5} \right) + c$ , where, $c$ is the constant of integration.

Note: This question can also be done directly using the integration formula after completing the square$\int {\sqrt {{a^2} - {x^2}} dx} = \dfrac{{x\sqrt {{a^2} - {x^2}} }}{2} + \dfrac{{{a^2}}}{2}\arcsin \left( {\dfrac{x}{a}} \right) + c$ .
Direct integration ${\cos ^2}x$ is not possible, so we have to convert it using the half-angle formula to remove the square.
Substituting the values back, for substitution is necessary, otherwise, the variables in the final answer won’t be the same as in the question.