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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- \frac{2}{3} + \frac{π}{4}

Decimal

0.11873 \dots

Solution steps

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

Rewrite using trig identities

= \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}

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