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

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

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

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

6 (- \frac{4}{3} + \frac{π}{2})

Decimal Notation

1.42477 \dots

Solution steps

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

Apply Integral Reduction

= 6 ([\frac{tan ^{3} ( x )}{3}]_{- \frac{π}{4}}^{\frac{π}{4}} - \int_{- \frac{π}{4}}^{\frac{π}{4}} tan^{2} (x) d x)

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

= 6 ([\frac{tan ^{3} ( x )}{3}]_{- \frac{π}{4}}^{\frac{π}{4}} - (- \frac{π}{2} + 2))

Simplify

6 ([\frac{tan ^{3} ( x )}{3}]_{- \frac{π}{4}}^{\frac{π}{4}} - (- \frac{π}{2} + 2)) : 6 ([\frac{1}{3} tan^{3} (x)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{π}{2} - 2)

= 6 ([\frac{1}{3} tan^{3} (x)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{π}{2} - 2)

Compute the boundaries:

\frac{2}{3}

= 6 (\frac{2}{3} + \frac{π}{2} - 2)

Simplify

= 6 (- \frac{4}{3} + \frac{π}{2})

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