Multiple answers to $\int \sqrt{4t - t^2} \, \textrm{d} t$ I'm trying to understand why I'm getting different answers when taking different approaches to integrating
$$
\int \sqrt{4t - t^2} \, \textrm{d} t
$$
First, I tried substituting $\sqrt t = 2 \sin \theta$:
$$
\begin{eqnarray}
\int \sqrt{t}\sqrt{4 - t} \, \textrm{d} t
&=& \int 2 \sin \theta \cdot \sqrt{4 - 4 \sin^2 \theta} \cdot 8 \sin \theta \cos \theta \, \textrm{d} \theta \\
&=& 32 \int \sin^2 \theta \cos^2 \theta \, \textrm{d} \theta \\
&=& 4 \int 1 - \cos 4 \theta \, \textrm{d} \theta \\
&=& 4 \theta - \sin 4 \theta + C \\
&=& 4 \theta - 4 \sin \theta \cos^3 \theta + 4 \sin^3 \theta \cos \theta + C \\
&=& 4 \arcsin \frac{\sqrt{t}}{2} + \frac{1}{2} (t - 2)\sqrt{4t - t^2} + C \\
\end{eqnarray}
$$
Second, I tried completing the square and substituting $t - 2 = 2 \sin \theta$:
$$
\begin{eqnarray}
\int \sqrt{4 - (t^2 - 4t + 4)} \, \textrm{d} t
&=& \int \sqrt{4 - (t - 2)^2} \, \textrm{d} t \\
&=& \int \sqrt{4 - 4 \sin^2 \theta} \cdot 2 \cos \theta \, \textrm{d} \theta \\
&=& 4 \int \cos^2 \theta \, \textrm{d} \theta \\
&=& 2 \int 1 - \cos 2 \theta \, \textrm{d} \theta \\
&=& 2 \theta + \sin 2 \theta + C \\
&=& 2 \theta + 2 \sin \theta \cos \theta + C \\
&=& 2 \arcsin \left(\frac{t - 2}{2}\right) + \frac{1}{2}(t - 2)\sqrt{4t - t^2} + C \\
\end{eqnarray}
$$
The second answer is the same as in the book but I don't understand why the first approach gives the wrong answer.
 A: The answers differ by a constant. In fact,
$$\arcsin\left(\frac t2-1\right) = 2\left(\arcsin\frac{\sqrt t}2 -\frac\pi4\right).\tag{*}$$
Proof: Write $\theta:=\arcsin\frac{\sqrt t}2$. Then $\sin^2\theta=\frac t4$ and $\cos^2\theta=1-\frac t4$, and using the angle-difference identities,
$$
\sin\left(\theta-\frac\pi4\right)=\frac1{\sqrt 2}(\sin\theta-\cos\theta)\tag1
$$
while
$$\cos\left(\theta-\frac\pi4\right)=\frac1{\sqrt 2}(\sin\theta+\cos\theta).\tag2$$
Therefore
$$
\sin2\left(\theta-\frac\pi4\right)=2\sin\left(\theta-\frac\pi4\right)\cos\left(\theta-\frac\pi4\right)
\stackrel{(1),(2)}=\sin^2\theta-\cos^2\theta =\frac t4-\left(1-\frac t4\right)=\frac t2-1.$$
Therefore both sides of (*) have the same sine. Similarly you can show that both sides have the same cosine.
A: If $$\sin\theta = \dfrac{\sqrt t}2,$$ then
$$\sin\left(\dfrac\pi2-2\theta\right) = \cos2\theta = 1-2\sin^2\theta = \dfrac{2-t}2,$$
$$2\theta = \dfrac\pi2-\arcsin\dfrac{2-t}2 = \dfrac\pi2+\arcsin\dfrac{t-2}2,$$
$$\theta = \dfrac\pi4+\dfrac12\arcsin\dfrac{t-2}2,$$
$$\boxed{\arcsin\dfrac{\sqrt t}2 = \dfrac12\arcsin\dfrac{t-2}2+\dfrac\pi4}$$
So the first answer is correct too
