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Evaluate the integral: integral from 0 to sqrt(2) of (x^2)/(sqrt(4 - x^2)) dx.

Question:

Evaluate the integral: {eq}\int_{0}^{\sqrt{2}} \frac{x^2}{\sqrt{4 - x^2}} \, \mathrm{d}x {/eq}.

Solving Integration using the concept of Substitution:

Let us consider the integral {eq}I = \int {f(g(x))} g'(x)dx {/eq}, where {eq}f(x) {/eq} and {eq}g(x) {/eq} are real valued functions of {eq}x {/eq}.

Let us assume the substitution {eq}z = g(x) \Rightarrow dz = g'(x)dx {/eq}

After substitution:

{eq}I = \int {f(z)} dz {/eq}, which is easily solvable.

Answer and Explanation:

Here the given integral is: {eq}\displaystyle \int_{0}^{\sqrt{2}} \frac{x^2}{\sqrt{4 - x^2}} \, \mathrm{d}x {/eq}

Solving the integration by taking the following substitution:

{eq}x = 2\sin \left( z \right) \Rightarrow dx = 2\cos \left( z \right)dz \\ \displaystyle x:0 \to \sqrt 2 \Rightarrow z:0 \to \displaystyle \frac{\pi }{4} \\ {/eq}

We get:

{eq}\begin{align*} \displaystyle I& = \int_0^{\sqrt 2 } {\frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}} {\mkern 1mu} {\rm{d}}x \\ \displaystyle & = \int_0^{\displaystyle \frac{\pi }{4}} {\frac{{{{\sin }^2}z}}{{2\cos z}}} {\mkern 1mu} 2\cos zdz \\ \displaystyle & = \int_0^{\displaystyle \frac{\pi }{4}} {{{\sin }^2}zdz} \\ \displaystyle & = \frac{1}{2}\int_0^{\displaystyle \frac{\pi }{4}} {\left( {2{{\sin }^2}z} \right)dz} \\ \displaystyle & = \frac{1}{2}\int_0^{\displaystyle \frac{\pi }{4}} {\left( {1 - \cos 2z} \right)dz} \\ \displaystyle & = \frac{1}{2}\left( {z - \frac{{\sin 2z}}{2}} \right)_0^{\frac{\pi }{4}} \\ \displaystyle & = \frac{1}{2}\left( {\frac{\pi }{4} - \frac{1}{2}} \right) \\ \end{align*} {/eq}


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