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Hola Yani
Creo que la tienes mal copiada:
$$\begin{align}&\sin \theta= \frac {2x}1=2x\\&\\&x= \frac 1 2 \sin \theta\\&\\&dx= \frac 1 2 \cos \theta d \theta\\&\\&\int \sqrt{1-4x^2}dx= \int \sqrt{1 -4( \frac 1 2 \sin \theta)^2} ·\frac 1 2 \cos \theta d \theta=\\&\\&\frac 1 2 \int \sqrt { 1 - \sin^2 \theta} \cos \theta d \theta=\\&\\&\frac 1 2 \int \cos \theta · \cos \theta d \theta= \frac 1 2 \int \cos^2 \theta d \theta==\\&\\&identidadº trigonométrica:\\&\cos^2 \theta= \frac{1 + \cos 2 \theta } 2\\&\\&== \frac 1 2 \int \frac 1 2 (1 +\cos 2 \theta )d \theta = \frac 1 4 \int 1+\cos 2 \theta d \theta =\\&\\&\frac 1 4\Big[ \theta + \frac 1 2 \sin 2 \theta \Big]==\\&\\&sin2 \theta =2sin \theta \cos \theta =2(2x)\sqrt{1-4x^2}\\&\\&== \frac 1 4 \Bigg[arcsin(2x)+\frac 1 2 2(2x)\sqrt{1-4x^2} \Bigg]=\\&\\& \frac 1 4 \Bigg[arcsin(2x)+(2x)\sqrt{1-4x^2} \Bigg]+C\\&\end{align}$$
Saludos
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