Hi, I’m a bit confused on how to solve a problem like this.

My friend helped me a bit, but didn’t explain very clearly, so I was wondering if you guys could help me understand this.

_{0}∫^{1/6}1/√(1-9x^{2})dx

So I was told that u=3x, and du=3

So it would become_{0}∫^{1/6}1/√(1-(3x)^{2}dx

Now for u-substitution.

1/√(1-u^{2})du

I get [arcsin(3x)]^{1/6}_{0}

Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:

(1/3)_{0}∫^{1/6}(1/√(1-(3x)^{2}) (3)dx

So yeah… there’s a 3 outside…

And there’s a 1/3 in front of the integration.

Can someone explain to me why this was done?

Also, can someone explained why u=3x, instead of something like u=9x?

Thanks J

Hi, I’m a bit confused on how to solve a problem like this.

My friend helped me a bit, but didn’t explain very clearly, so I was wondering if you guys could help me understand this.

_{0}∫^{1/6}1/√(1-9x^{2})dx

So I was told that u=3x, and du=3

So it would become_{0}∫^{1/6}1/√(1-(3x)^{2}dx

Now for u-substitution.

1/√(1-u^{2})du

I get [arcsin(3x)]^{1/6}_{0}

Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:

(1/3)_{0}∫^{1/6}(1/√(1-(3x)^{2}) (3)dx

So yeah… there’s a 3 outside…

And there’s a 1/3 in front of the integration.

Can someone explain to me why this was done?

Also, can someone explained why u=3x, instead of something like u=9x?

Thanks J