Question

Find the integration of the following function with respect to$x$ :$f(x) = \sqrt {2x - {x^2}}$

Answer 1

We have to solve the given integral using standard integration formulae but firstly we simplify the function into easier form in order to solve it. We convert the expression into standard algebraic expression using manipulation.

Complete solution step by step:
Firstly we write down the function given in the question:
$f\left( x \right) = \sqrt {2x - {x^2}}$
As we can see the degree of the expression is two so we change it into standard algebraic expression like this
$2x - {x^2} = 1 - 1 + 2x - {x^2} = 1 - \underbrace {\left( {1 - 2x + {x^2}} \right)}_{{\text{Expansion}}\,{\text{of}}\,{\text{standard}}\,{\text{identity}}} = 1 - {\left( {x - 1} \right)^2}$
Now we take the function as an integral with respect to $x$ like this
$I = \int {\sqrt {2x - {x^2}} } dx = \int {\sqrt {1 - {{\left( {x - 1} \right)}^2}} dx}$
So we take substitution method of integration and let
$s = x - 1, \Rightarrow ds = dx$
Putting these values into the above equation
$I = \int {\sqrt {1 - {{\left( {x - 1} \right)}^2}} } dx = \int {\sqrt {1 - {s^2}} } ds$
We know that $1 - {\sin ^2}x = {\cos ^2}x$ so we put $s = \sin \theta , \Rightarrow ds = \cos \theta d\theta$ these values into the above equation
$I = \int {\sqrt {1 - {s^2}} } ds = \int {\sqrt {1 - {{\sin }^2}\theta } } \cos \theta d\theta = \int {\cos \theta \times \cos \theta d\theta } \\\ I = \int {{{\cos }^2}\theta d\theta } \\\$
Using the following trigonometric identity and putting value
$\cos 2\theta = 2{\cos ^2}\theta - 1 \\\ \Rightarrow {\cos ^2}\theta = \dfrac{{\cos 2\theta + 1}}{2} \\\$
So we have
$I = \int {\left( {\dfrac{{\cos 2\theta + 1}}{2}} \right)d\theta = } \dfrac{1}{2}\int {\left( {\cos 2\theta + 1} \right)d\theta }$
We know that$\int {\cos \theta = \sin \theta }$. So we have
$I = \dfrac{1}{2}\left[ {\int {\cos 2\theta d\theta + \int {d\theta } } } \right] \\\ \Rightarrow I = \dfrac{1}{2}\left( {\dfrac{{\sin 2\theta }}{2} + \theta } \right) \\\ \Rightarrow I = \dfrac{{\sin 2\theta }}{4} + \dfrac{\theta }{2} \\\$
Now we use the trigonometric identity of $\sin 2\theta = 2\sin \theta \cos \theta$ and put it in the above equation
$I = \dfrac{{2\sin \theta \cos \theta }}{4} + \dfrac{\theta }{2} = \dfrac{{\sin \theta \cos \theta }}{2} + \dfrac{\theta }{2}$
We had already taken $s = \sin \theta , \Rightarrow {\sin ^{ - 1}}\left( s \right) = \theta$ and hence we have
$I = \dfrac{{s\sqrt {1 - {s^2}} }}{2} + \dfrac{{{{\sin }^{ - 1}}\left( s \right)}}{2}$
Now we substitute back the value of $s = x - 1$ in the equation
$I = \dfrac{{\left( {x - 1} \right)\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}{2} + \dfrac{{{{\sin }^{ - 1}}\left( {x \- 1} \right)}}{2} + C \\\ I = \dfrac{{\left( {x - 1} \right)\sqrt {2x - {x^2}} }}{2} + \dfrac{{{{\sin }^{ - 1}}\left( {x - 1} \right)}}{2} + C \\\$
Here $C$ is the integral constant which is obtained when we evaluate indefinite integral that is integration without limits.
Additional information: To find the exact or explicit solution we must apply integration limits and then we get the value of $C$ and after putting it in the integral value we get the final solution.

Note: We used the method of integration by substitution here in the question where we manipulate the expression by assuming a value of the given expression then we simplify it until we get a standard form to solve the expression by using integration formulae.