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

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

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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