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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.

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    $\begingroup$ Here's one sticking point: $1-\cos 4\theta = 8\sin^2\theta\cos^2\theta.$ $\endgroup$ – Cameron Williams Mar 2 '16 at 0:52
  • $\begingroup$ @CameronWilliams, good catch, I'll update my first answer. $\endgroup$ – Chewers Jingoist Mar 2 '16 at 0:56
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    $\begingroup$ Both answers are correct: if you let $f(t)=4\sin^{-1}\frac{\sqrt{t}}{2}-2\sin^{-1}\frac{t-2}{2}$, then $f^{\prime}(t)=0$ so $f(t)$ is a constant. $\endgroup$ – user84413 Mar 2 '16 at 1:08
  • $\begingroup$ @user84413, when I differentiate $f(t)$ I get $\frac{4}{\sqrt{4 - t}} - \frac{2}{\sqrt{4 - (t - 2)^2}} = \frac{4}{\sqrt{4 - t}} - \frac{2}{\sqrt{4t - t^2}}$. Could you elaborate on how you get $0$? $\endgroup$ – Chewers Jingoist Mar 2 '16 at 1:21
  • $\begingroup$ In the first term, I think you will get $4\frac{1}{\sqrt{1-t/4}}\frac{1}{4\sqrt{t}}=\frac{2}{\sqrt{4t-t^2}}$ $\endgroup$ – user84413 Mar 2 '16 at 1:33
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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.

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  • $\begingroup$ Where does $2$ come from at the start of your last line in $sin 2(\theta - \frac{\pi}{4})$? And what does the $(1),(2)$ mean? $\endgroup$ – Chewers Jingoist Mar 2 '16 at 1:31
  • $\begingroup$ They refer to the equations tagged (1) and (2). $\endgroup$ – grand_chat Mar 2 '16 at 1:32
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    $\begingroup$ @ChewersJingoist, the $2$ in $sin2(\theta - \frac{\pi}{4})$ comes from the identity $sin2x=2sinxcosx$. $\endgroup$ – Alexander Maru Mar 2 '16 at 1:39
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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

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