flyingpig
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Homework Statement
In my Calculus class, my professor told us that to handle integrals in the forms of \sqrt{a^2 - x^2}, we must use x = a\sin\theta. But my question is why? I tried it with x = acosθ and it works, but it just gives me an "opposite" answer.
Let's say \int \sqrt{9 - x^2} dx
Using x = 3sinθ, you should get
\frac{1}{2}\left ( \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )
The Attempt at a Solution
Now, let's try using x = 3cosθ
x = 3\cos\theta
dx = -3\sin\theta d\theta
\int \sqrt{9 - 9\cos^2 \theta } -3\sin\theta d\theta
-3\int \sqrt{9(1 - \cos^2 \theta)} \sin\theta d\theta
-9\int \sqrt{1 - \cos^2 \theta} \sin\theta d\theta
-9\int \sin^2\theta d\theta
-9\int \frac{1- \cos2\theta }{2} d\theta
-9\int \frac{1}{2}- \frac{\cos2\theta }{2} d\theta
-9\left [ \frac{\theta}{2}- \frac{\sin2\theta }{4} \right ]
-9\left [ \frac{\theta}{2}- \frac{2\sin\theta\cos\theta}{4} \right ]
Initially we had x = 3\cos\theta, so draw the triangle and we should get \sin\theta = \frac{\sqrt{3^2 - x^2}}{3} and \theta = \arccos\left ( \frac{x}{3} \right )
Our final answer should be
-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]
Now clearly
-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ] \neq \frac{1}{2}\left ( \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )
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