Problem:

Attachment 20622

I am stuck on this problem. Could it be as simple as changing the (tanx)^2 to (secx)^2

and having the answer as (cosx)^2 - x?

Or is it much more complicated than that? Please help me with this.

Printable View

- Jan 27th 2011, 10:23 PMflorxQuick integral question?
Problem:

Attachment 20622

I am stuck on this problem. Could it be as simple as changing the (tanx)^2 to (secx)^2

and having the answer as (cosx)^2 - x?

Or is it much more complicated than that? Please help me with this. - Jan 27th 2011, 10:39 PMLiverpool
$\displaystyle \dfrac{1-tan^2(x)}{sec^2(x)}=cos^2(x)-sin^2(x)=cos(2x)$

- Jan 28th 2011, 04:05 AMSoroban
Hello, florx!

What you're saying is*scary*. . .

Quote:

$\displaystyle \displaystyle \int\frac{1-\tan^2\!x}{\sec^2\!x}\,dx$

First of all: .$\displaystyle \tan^2\!x \;=\;\sec^2\!x - 1$

$\displaystyle \displaystyle\text{Substitute: }\;\frac{1 - (\sec^2\!x - 1)}{\sec^2\!x} \;=\;\frac{2-\sec^2\!x}{\sec^2\1x} \;=\; \frac{2}{\sec^2\!x} - \frac{\sec^2\!x}{\sec^2\!x}$

. . . . . . . $\displaystyle =\;2\cos^2\!x - 1 \;=\;\cos2x $ . as Liverpool suggested