Question

Use the trigonometric substitution \[\sqrt{x}=\sin \theta \] to determine \[\int\limits_{?}^{?}\sqrt{x}\sqrt{1-x}dx\]

Answer 1

If \(\sqrt x=\sin t\), then \(\dfrac{\mathrm{d}x}{2\sqrt x}=\cos t\,\mathrm{d}t\). From the second equation, you have \(\mathrm{d}x=2\sqrt x\cos t\,\mathrm{d}t=2\sin t\cos t\,\mathrm{d}t\).

Meanwhile, \(\sqrt x=\sin t\) means that for valid \(t\),
\[x=\sin^2t\implies 1-x=1-\sin^2t=\cos^2t\implies \sqrt{1-x}=\cos t\]
So the integral becomes
\[\int\color{red}{\sqrt x}\color{blue}{\sqrt{1-x}}\,\color{green}{\mathrm{d}x}=\int\color{red}{\sin t}\color{blue}{\cos t}\color{green}{2\sin t\cos t\,\mathrm{d}t}=2\int\sin^2t\cos^2t\,\mathrm{d}t\]

Answer 2

Oh ok, and then we can rewrite the integral using the half angle formulas and then take the antiderivative?

Answer 3

I got this, not sure how I would substitute back in the original variables to get the final answer.

\[2\int\limits_{?}^{?}\sin^2tcos^2tdt =\frac{ 1 }{ 4 }(t-\frac{ \sin(4t) }{ 4 })+C\]

Answer 4

\[\sin 4t = 2 \sin 2t \cos 2t = 2(2 \sin t \cos t)(1-2\sin^2 t)\]
\[\rightarrow \sin 4t = 4 \sqrt{x} \sqrt{1-x} (1-2x)\]