What is the integral from 0 to pi/4 of tan^2(θ) ?

The integral from 0 to pi/4 of tan^2(θ) is -pi/4+1

integral from 0 to pi/4 of tan^2(θ)

\int_{0}^{\frac{π}{4}} tan^{2} (θ) d θ

- \frac{π}{4} + 1

Decimal

0.21460 \dots

Solution steps

\int_{0}^{\frac{π}{4}} tan^{2} (θ) d θ

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} - 1 + sec^{2} (θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{\frac{π}{4}} 1 d θ + \int_{0}^{\frac{π}{4}} sec^{2} (θ) d θ

\int_{0}^{\frac{π}{4}} 1 d θ = \frac{π}{4}

\int_{0}^{\frac{π}{4}} sec^{2} (θ) d θ = 1

= - \frac{π}{4} + 1

\int_{0}^{e - 1} ln (x + 1) d x

\int_{0}^{1} \frac{x}{1 + x} d x

\int_{0}^{2} x^{2} - 1 d x

\int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

\int_{0}^{\frac{π}{4}} (sec (t))^{2} d t

What is the integral from 0 to pi/4 of tan^2(θ) ?
The integral from 0 to pi/4 of tan^2(θ) is -pi/4+1