integral from-1 to 0 of 1/(1+cos(t))

Lösung

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

tan (\frac{1}{2})

Dezimale

0.54630 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Integral einer Konstanten :

\int a d x = a x

= [1 \cdot u]_{- t a n (\frac{1}{2})}^{0}

Vereinfache

= [u]_{- t a n (\frac{1}{2})}^{0}

Berechne die Grenzen:

tan (\frac{1}{2})

= tan (\frac{1}{2})

\int_{1}^{2} (10 x^{4} - 9 x^{2}) d x

\int_{100}^{200} 5 + \frac{3}{x + 2} d x

\int_{1}^{e} 3 t^{2} ln (t) d t

\int_{- 2}^{3} (8 x - 4) d x

\int_{- 6 π}^{6 π} cos^{2} (x) d x