# How do you evaluate the definite integral int sec^2x/(1+tan^2x) from [0, pi/4]?

Aug 19, 2016

$\frac{\pi}{4}$

#### Explanation:

Notice that from the second Pythagorean identity that

$1 + {\tan}^{2} x = {\sec}^{2} x$

This means the fraction is equal to 1 and this leaves us the rather simple integral of

${\int}_{0}^{\frac{\pi}{4}} \mathrm{dx} = x {|}_{0}^{\frac{\pi}{4}} = \frac{\pi}{4}$

Aug 20, 2016

$\frac{\pi}{4}$

#### Explanation:

Interestingly enough, we can also note that this fits the form of the arctangent integral, namely:

$\int \frac{1}{1 + {u}^{2}} \mathrm{du} = \arctan \left(u\right)$

Here, if $u = \tan x$ then $\mathrm{du} = {\sec}^{2} x \mathrm{dx}$, then:

$\int {\sec}^{2} \frac{x}{1 + {\tan}^{2} x} \mathrm{dx} = \int \frac{1}{1 + {u}^{2}} \mathrm{du} = \arctan \left(u\right) = \arctan \left(\tan x\right) = x$

${\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} \frac{x}{1 + {\tan}^{2} x} \mathrm{dx} = {\left[x\right]}_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$