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

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

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

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

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

Decimal

0.15342 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{3} ( x )} d x

Apply u-substitution

: \int_{1}^{\frac{2}{2}} - \frac{1 - u ^{2}}{u ^{3}} d u

= \int_{1}^{\frac{2}{2}} - \frac{1 - u ^{2}}{u ^{3}} d u

Expand

- \frac{1 - u ^{2}}{u ^{3}} : - \frac{1}{u ^{3}} + \frac{1}{u}

= \int_{1}^{\frac{2}{2}} - \frac{1}{u ^{3}} + \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{2}{2}}^{1} - \frac{1}{u ^{3}} + \frac{1}{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_{\frac{2}{2}}^{1} \frac{1}{u ^{3}} d u + \int_{\frac{2}{2}}^{1} \frac{1}{u} d u)

\int_{\frac{2}{2}}^{1} \frac{1}{u ^{3}} d u = \frac{1}{2}

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

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

Simplify

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

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