What is the integral from 0 to pi/3 of tan^5(x) ?

The integral from 0 to pi/3 of tan^5(x) is ln(2)+3/4

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integral from 0 to pi/3 of tan^5(x)

\int_{0}^{\frac{π}{3}} tan^{5} (x) d x

ln (2) + \frac{3}{4}

Solution steps

\int_{0}^{\frac{π}{3}} tan^{5} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{3}} (- 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 = ln (2)

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

\int_{1}^{2} u^{3} d u = \frac{15}{4}

= ln (2) - 3 + \frac{15}{4}

Combine the fractions

- 3 + \frac{15}{4} : \frac{3}{4}

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

y^{^{''}} + \frac{6}{x} y^{^{'}} + \frac{4}{x ^{2}} y = 0

f (x) = x^{2} - 4 x + 2

x \to 1 lim (e^{(x^{3} - x)})

\frac{d}{d x} (e^{π x})

\frac{d}{d x} (\frac{1 + 2 y ^{2}}{y sin ( x )})

What is the integral from 0 to pi/3 of tan^5(x) ?
The integral from 0 to pi/3 of tan^5(x) is ln(2)+3/4