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$\int \sqrt{3-2x-x^2}\,dx$

First I did:

$$\begin{align}\int \sqrt{3-2x-x^2}\,dx &= \int \sqrt{-((x+1)^2-4)}\,dx \\ &=\int \sqrt{4-(x+1)^2}\,dx \\ \end{align}$$

Then I set $(x+1)=2\sin(t)$, $dx = 2\cos(t)\,dt$

$$\begin{align} \int \sqrt{4-(2\sin(t))^2}2\cos(t)\,dt &= \int\sqrt{4-4\sin^2(t)}\,\,2\cos(t)\,dt \\ &= \int4\sqrt{1-\sin^2(t)}\cos(t)\,dt \\ &=4\int \sqrt{1-\sin^2(t)}\cos(t)\,dt \\ &=4\int\cos^2(t)\,dt \\ &=4\int\frac{1+\cos(2t)}\,2\,dt \\ &=2 \int 1+\cos(2t) \,dt \\ &=2t + \sin(2t) + C \end{align}$$

$x+1 = 2\sin(t) \Leftrightarrow \frac{x+1}{2} = \sin(t) \Leftrightarrow t = \arcsin(\frac{x+1}{2})$

Replacing,

$$2\arcsin\left(\frac{x+1}{2}\right) + \sin\left(2\arcsin\left(\frac{x+1}{2}\right)\right) + C$$

But my book's solution is $$\frac{x+1}{2}\sqrt{3-2x-x^2}+2\arcsin\left(\frac{x+1}{2}\right)$$

So... what went wrong?

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2 Answers 2

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Hint:

$$\sin\left(2\arcsin\left(\dfrac u2\right)\right) = u\sqrt{1 - \dfrac{u^2}4}$$

In your case, $u = x + 1$. This may be helpful.

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HINT:

Note that $\sin(2x)=2\sin(x)\cos(x)$, $\sin(\arcsin(x))=x$, and $\cos(\arcsin(x))=\sqrt{1-x^2}$.

Can you reconcile now?

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