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

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

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integral from 0 to pi/4 of tan^3(θ)

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

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

Decimal Notation

0.15342 \dots

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \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

= - \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}

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

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