integral from 0 to 1 of pi(tan(pi/4 y))^2

Lösung

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

- π + 4

Dezimale

0.85840 \dots

Schritte zur Lösung

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{1} (tan (\frac{π}{4} y))^{2} d y

Vereinfache

= π \cdot \int_{0}^{1} tan^{2} (\frac{π}{4} y) d y

Wende U-Substitution an

= π \cdot \int_{0}^{\frac{π}{4}} tan^{2} (u) \frac{4}{π} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \frac{4}{π} \cdot \int_{0}^{\frac{π}{4}} tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= π \frac{4}{π} \cdot \int_{0}^{\frac{π}{4}} - 1 + sec^{2} (u) 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

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

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

\int_{0}^{\frac{π}{4}} sec^{2} (u) d u = 1

= π \frac{4}{π} (- \frac{π}{4} + 1)

Vereinfache

π \frac{4}{π} (- \frac{π}{4} + 1) : - π + 4

= - π + 4

\frac{\partial}{\partial x} (\frac{5 x + 4 y}{3 x + 2 y})

\int (tan (x))^{2} sec (x) d x

\frac{\partial}{\partial x} (6 - x^{3} - y^{5})

\frac{d}{d x} (arcsec (7 x))

d er i v a t i v e \frac{2 x ^{2} + 4 x}{( x + 1 ) ^{2}}