integral from 0 to pi/4 of tan^6(x)

Lösung

\int_{0}^{\frac{π}{4}} tan^{6} (x) d x

\frac{13}{15} - \frac{π}{4}

Dezimale

0.08126 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} tan^{6} (x) d x

Wende U-Substitution an

= \int_{0}^{1} \frac{u ^{6}}{1 + u ^{2}} d u

Lange Division

\frac{u ^{6}}{1 + u ^{2}} : u^{4} - u^{2} + 1 - \frac{1}{u ^{2} + 1}

= \int_{0}^{1} u^{4} - u^{2} + 1 - \frac{1}{u ^{2} + 1} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} u^{4} d u - \int_{0}^{1} u^{2} d u + \int_{0}^{1} 1 d u - \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

\int_{0}^{1} u^{4} d u = \frac{1}{5}

\int_{0}^{1} u^{2} d u = \frac{1}{3}

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} \frac{1}{u ^{2} + 1} d u = \frac{π}{4}

= \frac{1}{5} - \frac{1}{3} + 1 - \frac{π}{4}

Vereinfache

\frac{1}{5} - \frac{1}{3} + 1 - \frac{π}{4} : \frac{13}{15} - \frac{π}{4}

= \frac{13}{15} - \frac{π}{4}

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