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wrote...
2 weeks ago
What is the integral of √(a-y) /√y
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Educator
2 weeks ago
Is it with respect to y or a?

Let's assume it's with respect to y, the answer is:

\(\sqrt{a}\sqrt{y}\sqrt{1-\frac{y}{a}}+a\arcsin \left(\frac{\sqrt{y}}{\sqrt{a}}\right)\)

Here's how, you have to perform trigonometric substitution:

First set u = √y

\(2\int \sqrt{a-u^2}du\)

Then set \(u=\sqrt{a}\sin \left(v\right)\rightarrow v=\sin ^{-1}\left(\frac{u}{\sqrt{a}}\right),\ \frac{du}{dv}=\sqrt{a}\cos \left(v\right)\)

Then solve for du.

\(=\int \sqrt{a}\cos \left(v\right)\cdot \sqrt{a-a\sin ^2\left(v\right)}dv\)

Simplify:

\(a-a\sin ^2\left(v\right)=a\cos ^2\left(v\right)\)

\(=a\int \cos ^2\left(v\right)dv\)

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