# Quick integral question?

• January 27th 2011, 11:23 PM
florx
Quick 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?

• January 27th 2011, 11:39 PM
Liverpool
$\dfrac{1-tan^2(x)}{sec^2(x)}=cos^2(x)-sin^2(x)=cos(2x)$
• January 28th 2011, 05:05 AM
Soroban
Hello, florx!

What you're saying is scary . . .

Quote:

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

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

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

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