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

The integral from 0 to pi/4 of (tan^2(x)+2x) is -pi/4+1+(pi^2)/(16)

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

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

- \frac{π}{4} + 1 + \frac{π ^{2}}{16}

Decimal

0.83145 \dots

Solution steps

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

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

\int_{0}^{\frac{π}{4}} 2 x d x = \frac{π ^{2}}{16}

= - \frac{π}{4} + 1 + \frac{π ^{2}}{16}

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What is the integral from 0 to pi/4 of (tan^2(x)+2x) ?
The integral from 0 to pi/4 of (tan^2(x)+2x) is -pi/4+1+(pi^2)/(16)