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

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

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

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

- \frac{2}{3} + \frac{π}{4}

Decimal Notation

0.11873 \dots

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

= [\frac{1}{3} tan^{3} (x)]_{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}

y^{^{''}} - 2 y^{^{'}} + y = e^{2 x}

\int_{0}^{l n (19)} x e^{x} d x

\int t sin (n t) d t

\frac{\partial}{\partial x} (cot (x))

\frac{5}{x}

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