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

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}

\frac{d}{d x} (a - b x)

\int - \frac{sin ( x )}{cos ^{3} ( x )} d x

\frac{d}{d x} ((sin (x) + x)^{2} + 7 cos (5 x))

t an g e n t - \frac{8 x}{x ^{2} + 1}

\frac{d}{d x} (x^{\frac{2}{3}} \cdot (x - 4)^{2})

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