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

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

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

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

2 (- \frac{2}{3} + \frac{π}{4})

Decimal

0.23746 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integral Reduction

= 2 ([\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

= 2 ([\frac{tan ^{3} ( x )}{3}]_{0}^{\frac{π}{4}} - (- \frac{π}{4} + 1))

Simplify

2 ([\frac{tan ^{3} ( x )}{3}]_{0}^{\frac{π}{4}} - (- \frac{π}{4} + 1)) : 2 ([\frac{1}{3} tan^{3} (x)]_{0}^{\frac{π}{4}} + \frac{π}{4} - 1)

= 2 ([\frac{1}{3} tan^{3} (x)]_{0}^{\frac{π}{4}} + \frac{π}{4} - 1)

Compute the boundaries:

\frac{1}{3}

= 2 (\frac{1}{3} + \frac{π}{4} - 1)

Simplify

= 2 (- \frac{2}{3} + \frac{π}{4})

5 sin (x)

f (x) = \frac{2}{x + 2}

4 tan (x)

\frac{d}{d x} (\frac{x - 3}{x + 3})

\frac{d}{d x} (\frac{- 2 x ^{2}}{5 y ^{2}})

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