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.

_{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

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.

Re: Inverse Trig Func: Integration

You have $\displaystyle \displaystyle \begin{align*} \int_0^{\frac{1}{6}}{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int_0^{\frac{1}{6}}{\frac{1}{\sqrt{1 - \left( 3x \right) ^2 } } \,dx} \end{align*}$. The standard integral is $\displaystyle \displaystyle \begin{align*} \int{\frac{1}{\sqrt{1 - X^2}}\,dX} = \arcsin{(X)} + C \end{align*}$, so the substitution $\displaystyle \displaystyle \begin{align*} u = 3x \end{align*}$ here is appropriate to turn it into a multiple of this standard integral.

Re: Inverse Trig Func: Integration

Oh, now I see where it is coming from!

Thanks you guys! :)