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I'm studying for my finals and I have this integral that I'm trying to evaluate (part of a bigger problem):

$\int\sqrt{150^2-x^2} \cdot dx$

I have evaluated a few integrals of this type before so the first thought that came to my mind was to substitute $x = \sin t$ and $dx = \cos t \cdot dt$.

So now I have:

$\int \sqrt{150^2-\sin^2t} \cdot \cos t \cdot dt$

However, here is where I'm getting stuck. Usually instead of having $150^2$ I have $1$, and by using $1-\sin^2t = \cos^2t \space$ I can continue, but not in this case.

How should I go on?

2 Answers 2

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With such integration by parts problems where $\sqrt{a^2-x^2}$ appears and a trigonometric substitution is appropriate, the substitution to make is $x=a\sin t$. Then we have $a^2(1-\sin^2 t)$ in the square root, which simplifies nicely.

Likewise one would substitute $x=a \tan t$ into the integral $\displaystyle \int \frac{1}{a^2+x^2}\, dx$, and so on.

For a more detailed explanation, see here (page 3).

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    Thanks! Solved it using $x = a sin t$. What I was doing before was basically the same, but because $a = 1$ our teacher never gave us the right was to substitute.2012-01-08
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What about doing first the change of variables $ y = x/150$, and then your change of variables ?

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    yes i do understand that, i just wanted the OP to know what really was going on. By the way, your answer, though neat, does lack a bit of description. It would've been better if you had elaborated a bit more. Maybe you had no time whilst writing down the answer but still, it would have avoided discussions like ours :)2012-01-09