# Inverse Trig Func: Integration

• Apr 27th 2013, 07:47 AM
Chaim
Inverse Trig Func: Integration
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.
01/6 1/√(1-9x2)dx
So I was told that u=3x, and du=3
So it would become 01/6 1/√(1-(3x)2dx
Now for u-substitution.
1/√(1-u2)du

I get [arcsin(3x)]1/60
Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:
(1/3)01/6 (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.
01/6 1/√(1-9x2)dx
So I was told that u=3x, and du=3
So it would become 01/6 1/√(1-(3x)2dx
Now for u-substitution.
1/√(1-u2)du

I get [arcsin(3x)]1/60
Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:
(1/3)01/6 (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
• Apr 27th 2013, 04:41 PM
chiro
Re: Inverse Trig Func: Integration
Hey Chaim.

The answer for the 3 and 1/3 is that du = 3dx.

This means that 1/3 * 3 dx = dx = 1/3 * du and hence if you are integrating with respect to u then you need a 1/3 term to balance everything out.
• Apr 27th 2013, 05:52 PM
Prove It
Re: Inverse Trig Func: Integration
You have \displaystyle \begin{align*} \int_0^{\frac{1}{6}}{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int_0^{\frac{1}{6}}{\frac{1}{\sqrt{1 - \left( 3x \right) ^2 } } \,dx} \end{align*}. The standard integral is \displaystyle \begin{align*} \int{\frac{1}{\sqrt{1 - X^2}}\,dX} = \arcsin{(X)} + C \end{align*}, so the substitution \displaystyle \begin{align*} u = 3x \end{align*} here is appropriate to turn it into a multiple of this standard integral.
• Apr 27th 2013, 07:09 PM
Chaim
Re: Inverse Trig Func: Integration
Oh, now I see where it is coming from!

Thanks you guys! :)