integral from 0 to 1 of sqrt(2-2cos(t))

Lösung

\int_{0}^{1} 2 - 2 cos (t) d t

2 (- \frac{2}{sec ( \frac{1}{2} )} + 2)

Dezimale

0.48966 \dots

Schritte zur Lösung

\int_{0}^{1} 2 - 2 cos (t) d t

Wende U-Substitution an

= \int_{0}^{t a n (\frac{1}{2})} \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{t a n (\frac{1}{2})} \frac{u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Wende U-Substitution an

= 4 \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{2 v v} d v

Entferne die Konstante:

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

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{v v} d v

Vereinfache

v v : v^{\frac{3}{2}}

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{v ^{\frac{3}{2}}} d v

Wende die Potenzregel an

= 4 \cdot \frac{1}{2} [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})}

Vereinfache

4 \cdot \frac{1}{2} [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})} : 2 [- \frac{2}{v}]_{1}^{s e c^{2} (\frac{1}{2})}

= 2 [- \frac{2}{v}]_{1}^{s e c^{2} (\frac{1}{2})}

Berechne die Grenzen:

- \frac{2}{sec ( \frac{1}{2} )} + 2

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

\int_{1}^{4} (x^{3}) d x

\int_{\frac{x}{2}}^{x} - sin (x) d x

\int_{1}^{2} \frac{4 x ^{2} - 3 x + 4}{x} d x

\int_{0}^{4} π (16 - x^{2})^{2} d x

\int_{- 1}^{3} (6 x y - 5 y) d y