What is the correct answer for this trigonometric improper integral?

[cmab]

Here's a integral where I have to use trigonometric substitution but I can't get the right answer.

[int a=0 b=3] 1/(sqrt[9-x^2]) dx

I did the limit as t approches 3 from the left.

Then i did my trigonometric substitution, and it gives me arcsin(x/3).

Then i computed what i had arcsin(a/3)-arcsin(0/3).

It gives me 1.57 (estimated) or Pie/2 (real)

But in the answer sheet, it says 9pie/4...

 

[Jameson]

Your integral is correct.

$$\int\frac{1}{\sqrt{9-x^2}}dx = \arcsin{\frac{x}{3}} + C$$

When you apply the bounds, you get $$\arcsin{1} - \arcsin{0}$$

The arcsin of 0 is 0 and the arcsin of 1 is $$\frac{\pi}{2}$$

I don't see anything wrong with your answer.

 

[cmab]

Your integral is correct.

$$\int\frac{1}{\sqrt{9-x^2}}dx = \arcsin{\frac{x}{3}} + C$$

When you apply the bounds, you get $$\arcsin{1} - \arcsin{0}$$

The arcsin of 0 is 0 and the arcsin of 1 is $$\frac{\pi}{2}$$

I don't see anything wrong with your answer.

I don't know man, maybe the answer sheet is wrong. It says 9pie/4,as I mentionned before.

I tried, I had pie/2, and test it on graphmatica the program, and it gave something near it.

 

[krab]

Well, either your answer sheet is wrong, or you've described the problem incorrectly.