What is the integral from 0 to pi/4 of tan^5(x) ?

The integral from 0 to pi/4 of tan^5(x) is -1/4+1/2 ln(2)

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integral from 0 to pi/4 of tan^5(x)

\int_{0}^{\frac{π}{4}} tan^{5} (x) d x

- \frac{1}{4} + \frac{1}{2} ln (2)

Decimal

0.09657 \dots

Solution steps

\int_{0}^{\frac{π}{4}} tan^{5} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

\int_{1}^{2} u^{3} d u = \frac{3}{4}

= \frac{1}{2} ln (2) - 1 + \frac{3}{4}

Combine the fractions

- 1 + \frac{3}{4} : - \frac{1}{4}

= - \frac{1}{4} + \frac{1}{2} ln (2)

\int_{0}^{π} 19 sin^{2} (4 x) d x

\int t + 4 d t

f (x) = sin (sin (x)), at x = 3 π, 0

a^{x}

(3 x^{2} y^{2} + y) d x + (2 x^{3} y + x) d y = 0

What is the integral from 0 to pi/4 of tan^5(x) ?
The integral from 0 to pi/4 of tan^5(x) is -1/4+1/2 ln(2)