What is the integral from 0 to pi/4 of 6tan^4(x) ?

The integral from 0 to pi/4 of 6tan^4(x) is 6(-2/3+pi/4)

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integral from 0 to pi/4 of 6tan^4(x)

\int_{0}^{\frac{π}{4}} 6 tan^{4} (x) d x

6 (- \frac{2}{3} + \frac{π}{4})

Decimal Notation

0.71238 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 6 tan^{4} (x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{\frac{π}{4}} tan^{4} (x) d x

Apply Integral Reduction

= 6 ([\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

= 6 ([\frac{tan ^{3} ( x )}{3}]_{0}^{\frac{π}{4}} - (- \frac{π}{4} + 1))

Simplify

6 ([\frac{tan ^{3} ( x )}{3}]_{0}^{\frac{π}{4}} - (- \frac{π}{4} + 1)) : 6 ([\frac{1}{3} tan^{3} (x)]_{0}^{\frac{π}{4}} + \frac{π}{4} - 1)

= 6 ([\frac{1}{3} tan^{3} (x)]_{0}^{\frac{π}{4}} + \frac{π}{4} - 1)

Compute the boundaries:

\frac{1}{3}

= 6 (\frac{1}{3} + \frac{π}{4} - 1)

Simplify

= 6 (- \frac{2}{3} + \frac{π}{4})

x \cdot arcsin (x)

f (x) = x e^{2 x} sin (6 x)

\frac{d}{d x} (ln (x^{2} - 1))

x \to 4 lim (x^{2} - 16 x - 4)

\int \frac{x}{5 + x ^{2}} d x

What is the integral from 0 to pi/4 of 6tan^4(x) ?
The integral from 0 to pi/4 of 6tan^4(x) is 6(-2/3+pi/4)