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

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

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integral from 0 to pi/4 of tan^2(x)

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

- \frac{π}{4} + 1

Decimal Notation

0.21460 \dots

Solution steps

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

Rewrite using trig identities

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

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

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

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

= - \frac{π}{4} + 1

n = 1 \sum \infty \frac{1}{5 ^{n + 3}}

\int \frac{3 x ^{2}}{125} d x

sin (cos (tan (x)))

\frac{d}{d x} (1 + cos (2 x))

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