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Lets say you must perform a simple integral with trigonometric substitution - we'll choose $\int{\sqrt{1-x^2}dx}$ from $0$ to $17$. Now we use $x = \sin(t)$ for our substitution. Here's my problem: the limits of integration for $x$ range from $0$ to $17$, but $\sin$ never achieves a value greater than $1$. How do we change the limits of integration to accommodate this?

Note: I am not interested in writing in the limits at the very end, after back-substituting.

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If you are dealing with real calculus $1-x^2\ge0\implies -1\le x\le1$ –  lab bhattacharjee Jul 14 '13 at 16:19
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You should not be trying to integrate the function from $0$ to $17$, it is not defined beyond $1$. –  André Nicolas Jul 14 '13 at 16:19
    
"How do we change the limits of integration to accommodate this?" Simple answer: you don't. –  Ataraxia Jul 14 '13 at 16:39
    
@AndréNicolas well it's just not defined in the real domain –  Wolter Hellmund Sep 8 '13 at 21:05
    
@WolterHellmund: The OP is likely doing a first calculus course. Complex integration can wait. –  André Nicolas Sep 8 '13 at 21:09

2 Answers 2

The function $\sqrt{1-x^2}$ is not defined for $x\gt 1$, so you will not be wanting to integrate it from $0$ to $17$. Thus the issue does not arise.

As $t$ travels from $-\pi$ to $\pi$, $\sin t$ travels smoothly from $-1$ to $1$, so the substitution is suitable for $\int_a ^b\sqrt{1-t^2}\,dt$, where $-1\le a\le b\le 1$.

Once can imagine running into problems, with for example $\int_0^3\sqrt{17-x^2}\,dx$. This is perfectly well defined. The substitution $x=\sin t$ would not be suitable, since $\sin t$ does not take on values in the interval $(2,3]$. Luckily, the substitution is not useful! For this problem we would use $x=\sqrt{17}\sin t$.

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If you are substituting $x=\sin t$, then the resultant integral should be $\displaystyle{\int_{0}^{\frac{\pi}{2}}{\cos^2 t dt}}$

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