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

The integral from 0 to pi/4 of tan^4(t) is -2/3+pi/4

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

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

- \frac{2}{3} + \frac{π}{4}

Decimal Notation

0.11873 \dots

Solution steps

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

Apply Integral Reduction

= [\frac{tan ^{3} ( t )}{3}]_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} tan^{2} (t) d t

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

= [\frac{tan ^{3} ( t )}{3}]_{0}^{\frac{π}{4}} - (- \frac{π}{4} + 1)

Simplify

[\frac{tan ^{3} ( t )}{3}]_{0}^{\frac{π}{4}} - (- \frac{π}{4} + 1) : [\frac{1}{3} tan^{3} (t)]_{0}^{\frac{π}{4}} + \frac{π}{4} - 1

= [\frac{1}{3} tan^{3} (t)]_{0}^{\frac{π}{4}} + \frac{π}{4} - 1

Compute the boundaries:

\frac{1}{3}

= \frac{1}{3} + \frac{π}{4} - 1

Simplify

= - \frac{2}{3} + \frac{π}{4}

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What is the integral from 0 to pi/4 of tan^4(t) ?
The integral from 0 to pi/4 of tan^4(t) is -2/3+pi/4