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

The integral from 0 to pi/4 of 6tan^3(x) is 6(-1/2 ln(2)+1/2)

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integral from 0 to pi/4 of 6tan^3(x)

\int_{0}^{\frac{π}{4}} 6 tan^{3} (x) d x

6 (- \frac{1}{2} ln (2) + \frac{1}{2})

Decimal Notation

0.92055 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 6 tan^{3} (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^{3} (x) d x

Rewrite using trig identities

= 6 \cdot \int_{0}^{\frac{π}{4}} (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

: \int_{1}^{2} \frac{- 1 + u ^{2}}{u} d u

= 6 \cdot \int_{1}^{2} \frac{- 1 + u ^{2}}{u} d u

Expand

\frac{- 1 + u ^{2}}{u} : - \frac{1}{u} + u

= 6 \cdot \int_{1}^{2} - \frac{1}{u} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (- \int_{1}^{2} \frac{1}{u} d u + \int_{1}^{2} u d u)

\int_{1}^{2} \frac{1}{u} d u = \frac{1}{2} ln (2)

\int_{1}^{2} u d u = \frac{1}{2}

= 6 (- \frac{1}{2} ln (2) + \frac{1}{2})

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

f (x) = (4 x^{2} - 5 x) e^{x}

\int 5 e^{- 4 x} d x

\frac{d}{d x} (arcsin (\frac{9}{x ^{2}}))

(\frac{4 x}{x ^{2} + 1})^{\frac{1}{4}}

What is the integral from 0 to pi/4 of 6tan^3(x) ?
The integral from 0 to pi/4 of 6tan^3(x) is 6(-1/2 ln(2)+1/2)