Inverse Trigonometry.

**Sunyata** · November 15th 2009, 08:11 PM

A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

What exactly am I meant to do for these types of questions?

**redsoxfan325** · November 15th 2009, 08:20 PM

Originally Posted by Sunyata

A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

What exactly am I meant to do for these types of questions?

To find the original function you need to integrate

$\int\frac{1-x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx-\int\frac{x}{1+x^2}\,dx$

The first integral is $\arctan(x)$ (you should find this on any table of integrals) and the second integral can be solved using a simple u-sub ( $u=1+x^2$ ), so you end up with $f(x)=\arctan(x)-\frac{1}{2}\ln(1+x^2)$ . Then you set $f(x)=0$ to get the answer.

It seems a little strange that the question would ignore the presence of the integration constant though.

Thread: Inverse Trigonometry.

LinkBack

Thread Tools

Display

Inverse Trigonometry.

Similar Math Help Forum Discussions

inverse trigonometry 2

inverse trigonometry

inverse trigonometry

Inverse Trigonometry

Inverse trigonometry?

Search Tags