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#### Sherlock

##### Member

- Jan 28, 2012

- 59

Consider $\displaystyle I = \displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}.$ It can be shown that this is $\pi$.

Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$

However, the latter integral is equal to $-\pi$. Where is the negative from?

Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$

However, the latter integral is equal to $-\pi$. Where is the negative from?

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