What is the integral from 0 to 1 of tan^3(x) ?

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

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integral from 0 to 1 of tan^3(x)

\int_{0}^{1} tan^{3} (x) d x

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

Decimal Notation

0.59713 \dots

Solution steps

\int_{0}^{1} tan^{3} (x) d x

Rewrite using trig identities

= \int_{0}^{1} (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

: \int_{1}^{s e c (1)} \frac{- 1 + u ^{2}}{u} d u

= \int_{1}^{s e c (1)} \frac{- 1 + u ^{2}}{u} d u

Expand

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

= \int_{1}^{s e c (1)} - \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}^{s e c (1)} \frac{1}{u} d u + \int_{1}^{s e c (1)} u d u

\int_{1}^{s e c (1)} \frac{1}{u} d u = ln (sec (1))

\int_{1}^{s e c (1)} u d u = \frac{sec ^{2} ( 1 ) - 1}{2}

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

\int_{2}^{5} \frac{1}{x ^{4}} + 10 x^{5} d x

\int_{0}^{1} \frac{e ^{2 x} - 1}{2} d x

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

\int_{0}^{π} - sin^{3} (x) d x

\int_{0}^{L} x \cdot sin (\frac{nπ x}{L}) d x

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