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I have the following definite integral:

$$\int^{5}_{-5}{\sqrt{25-x^2} dx}$$

I do a variable substitution:

$$y = 25 - x^2$$ $$x = \sqrt{25 - y}$$ $$dy = -2x~dx$$ $$dx = \frac{dy}{-2x} = -\frac{1}{2\sqrt{25-y}}$$

I get the new integral:

$$-\frac{1}{2}\int^{0}_{0}{\sqrt{\frac{y}{25-y}} dy}$$

From Integral table I get that this integral (24) may be solved as:

$$-\frac{1}{2}(-\sqrt{y(25-y)}-25\tan^{-1}\frac{\sqrt{y(25-y)}}{y-25})\biggr\rvert^0_{0}$$

Going back to $x$ I get:

$$-\frac{1}{2}(-\sqrt{(25-x^2)x^2}-25\tan^{-1}\frac{\sqrt{(25-x^2)x^2}}{-x^2})\biggr\rvert^5_{-5}$$

Then:

$$\frac{1}{2}x\sqrt{25-x^2}+\frac{25}{2}\tan^{-1}\frac{-\sqrt{25-x^2}}{x}\biggr\rvert^5_{-5}$$

But I get $0$ as result, while the right answer is $\frac{25\pi}{2}$.

Where is the mistake?

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    Use $x = 5\sin t$ instead. – 2017-02-23
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    Can't you see that is the area of a half-circle? – 2017-02-23
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    @JackD'Aurizio Yes. But I want to solve it algebraically. – 2017-02-24

2 Answers 2

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$\sqrt{25-x^2}$ is an even function. So $$\boxed{\int_{-5}^5 \sqrt{25-x^2} dx=2\cdot \int_{0}^5 \sqrt{25-x^2} dx}$$

Now proceed the way you have, by substitution with $y$ and so on. You will get the desired result.

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    Why the down votes? It is a far easy method. – 2017-02-23
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    I agree, because of the symmetry, the solution looks easy. I will up vote this. – 2017-02-23
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The "mistake" is $x=\sqrt{25 - y}$. What happens if $x < 0$? One needs to be careful when making non-injective substitutions.

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    Which substitution should I use to overcome this problem? – 2017-02-23