Evaluate the definite integral $\int_{-2}^0 \frac3{\sqrt{-x^2-2x}}\ dx.$

Question

Evaluate $$\int_{-2}^0 \frac3{\sqrt{-x^2-2x}}\ dx.$$

I have tried substituting the square root, but might have done the substitution wrong. I also end up with having to divide one by zero and don't really know what to do then (please see picture:[1]: https://i.stack.imgur.com/SCqgA.jpg). Have I reasoned completely wrong? I'm be super grateful for any help.

/Nicklas

I've written an answer; it requires you to know that $\arcsin x+C$ is the antiderivative of $\frac1{\sqrt{1-x^2}}$. — Andrew Chin, Dec 03 '19 at 15:22

score 2 · Accepted Answer · answered Dec 03 '19 at 15:21

2

We have \begin{align}\int_{-2}^0 \frac3{\sqrt{-x^2-2x}}\ dx&=3\int_{-2}^0\frac1{\sqrt{1-(x+1)^2}}\ dx\\ &=3\int_{-1}^1\frac{1}{\sqrt{1-u^2}}\ du\\ &=3\arcsin u\ \bigg|_{-1}^1\\ &=3\left(\arcsin 1-\arcsin\left(-1\right)\right)\\ &=3\left(\frac{\pi}2-\left(-\frac{\pi}2\right)\right)\\ &=\boxed{3\pi} \end{align}

answered Dec 03 '19 at 15:21

Andrew Chin

7,389

1

I just wrote the same +1 :-) – Math-fun Dec 03 '19 at 15:25

Evaluate the definite integral $\int_{-2}^0 \frac3{\sqrt{-x^2-2x}}\ dx.$

1 Answers1