Simple, yet ambiguous integral question

[TheExibo]

Homework Statement

Evaluate the definite (from -0.3 to 0.3) integral ∫tanxdx

Homework Equations

(dy/dx)tanx=(secx)^2

The Attempt at a Solution

Using the anti-derivative, I got to (sec(0.3))^2-(sec(-0.3))^2, which gives approximately 2. However, if the derivative is sketched out, and the area underneath is found, the answer is then 0.

I found this happening with other similar questions as well. What could be going on?

 

[ShayanJ]

$\frac{d}{dx} \tan x=\sec^2 x$ implies $\int \sec^2 x \ dx=\tan x$, not $\int \tan x \ dx=\sec^2 x$!
$\tan x$ is an odd function of x so its definite integral over -b to b, for any real b, is 0.
Of course, for some b, the interval may contain a singular point of $\tan x$ in which case the definite integral would diverge.

 

[Ray Vickson]

Homework Statement

Evaluate the definite (from -0.3 to 0.3) integral ∫tanxdx

Homework Equations

(dy/dx)tanx=(secx)^2

The Attempt at a Solution

Using the anti-derivative, I got to (sec(0.3))^2-(sec(-0.3))^2, which gives approximately 2. However, if the derivative is sketched out, and the area underneath is found, the answer is then 0.

I found this happening with other similar questions as well. What could be going on?

You have the wrong antiderivative, but that is not your only error: you should get an answer of 0 (not $\approx$ 2) because $\sec^2(0.3) = \sec^2(-0.3)$.