Advanced Integration Problem

[Jason76]

\(\displaystyle \dfrac{dy}{dx} = \dfrac{x}{\sqrt{4x - x^{2}}}\)

\(\displaystyle \int dy = \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

\(\displaystyle y = \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

\(\displaystyle \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

Converting to form:

\(\displaystyle \int \dfrac{du}{\sqrt{2au - u^{2}}} = \sin^{-1} \dfrac{u - a}{a}\) and \(\displaystyle \int u^{n} du = \dfrac{n^{n + 1}}{n + 1}\)

\(\displaystyle \int \dfrac{x}{\sqrt{4x - x^{2}}} dx = \dfrac{[2 - (2 -x)]}{\sqrt{4x - x^{2}}}dx\) What is going on here.

\(\displaystyle 2 \int \dfrac{dx}{\sqrt{4x - x^{2}}} - \int \dfrac{(2 - x) dx}{\sqrt{4x - x^{2}}}\)

\(\displaystyle 2 \int \dfrac{2 dx}{\sqrt{4x - x^{2}}} - \int (4x - x^{2})^{-1/2} (2 - x) dx\)

\(\displaystyle u = 4x - x^{2}\)

\(\displaystyle du = 4 - 2x dx\)

\(\displaystyle \dfrac{1}{2}du = 2 - x dx\)

\(\displaystyle 2 \int \dfrac{2 dx}{\sqrt{4x - x^{2}}} - \dfrac{1}{2} \int (4x - x^{2})^{-1/2} (2 - x) dx\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - (\dfrac{1}{2}) \dfrac{u^{1/2}}{\dfrac{1}{2}} + C\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - (\dfrac{1}{2})(2) u^{1/2} + C\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - u^{1/2} + C\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - (4x - x^{2})^{1/2} + C\)

 

[stapel]

\(\displaystyle \dfrac{dy}{dx} = \dfrac{x}{\sqrt{4x - x^{2}}}\)

\(\displaystyle \int dy = \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

\(\displaystyle y = \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

\(\displaystyle \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

Converting to form:

\(\displaystyle \int \dfrac{du}{\sqrt{2au - u^{2}}} = \sin^{-1} \dfrac{u - a}{a}\) and \(\displaystyle \int u^{n} du = \dfrac{n^{n + 1}}{n + 1}\)

\(\displaystyle \int \dfrac{x}{\sqrt{4x - x^{2}}} dx = \dfrac{[2 - (2 -x)]}{\sqrt{4x - x^{2}}}dx\) What is going on here.

Don't you know what you are doing in your working? :shock:

 

[Jason76]

I am copying what is in the book. But I understand everything but the lines I mentioned.

 

[Deleted member 4993]

\(\displaystyle \dfrac{dy}{dx} = \dfrac{x}{\sqrt{4x - x^{2}}}\)

\(\displaystyle \int dy = \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

\(\displaystyle y = \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

\(\displaystyle \int \dfrac{x}{\sqrt{4x - x^{2}}} dx\)

Converting to form:

\(\displaystyle \int \dfrac{du}{\sqrt{2au - u^{2}}} = \sin^{-1} \dfrac{u - a}{a}\) and \(\displaystyle \int u^{n} du = \dfrac{n^{n + 1}}{n + 1}\)

\(\displaystyle \int \dfrac{x}{\sqrt{4x - x^{2}}} dx = \dfrac{[2 - (2 -x)]}{\sqrt{4x - x^{2}}}dx\) What is going on here. - Right!!! Where did the integration symbol \(\displaystyle \int\) go?

Other than that simple algebra!!! 2 - (2 - x) = 2 - 2 + x = x

\(\displaystyle 2 \int \dfrac{dx}{\sqrt{4x - x^{2}}} - \int \dfrac{(2 - x) dx}{\sqrt{4x - x^{2}}}\)

\(\displaystyle 2 \int \dfrac{2 dx}{\sqrt{4x - x^{2}}} - \int (4x - x^{2})^{-1/2} (2 - x) dx\)

\(\displaystyle u = 4x - x^{2}\)

\(\displaystyle du = 4 - 2x dx\)

\(\displaystyle \dfrac{1}{2}du = 2 - x dx\)

\(\displaystyle 2 \int \dfrac{2 dx}{\sqrt{4x - x^{2}}} - \dfrac{1}{2} \int (4x - x^{2})^{-1/2} (2 - x) dx\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - (\dfrac{1}{2}) \dfrac{u^{1/2}}{\dfrac{1}{2}} + C\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - (\dfrac{1}{2})(2) u^{1/2} + C\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - u^{1/2} + C\)

\(\displaystyle \rightarrow 2 \sin^{-1} \dfrac{x - 2}{2} - (4x - x^{2})^{1/2} + C\)

.

 

[stapel]

I am copying what is in the book.

And the book started the section with no text, no words, no title, no explanation, nothing? It started with the first line of your post, with no explanation of what might be going on? I rather doubt that.

Seriously, we really can't read your mind. :shock:

 

[Jason76]

I see what's going on. This is some kind of De-composition thing, regarding the lines I pointed out.