integral from 0 to pi/2 of 1/(3+5cos(x))

Lösung

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 5 cos ( x )} d x

\frac{ln ( 3 )}{4}

Dezimale

0.27465 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 5 cos ( x )} d x

Wende U-Substitution an

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

Wende integrale Substitution an

= \int_{0}^{\frac{1}{2}} \frac{1}{2 ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} \frac{1}{- v ^{2} + 1} d v

Nutze das gemeinsame Integral :

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Vereinfache

\frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}} : \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}}

= \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Berechne die Grenzen:

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

= \frac{1}{2} \cdot \frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

Vereinfache

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

\int_{4}^{5} (x - 4) - (x - 4)^{3} d x

\int_{1}^{2} \frac{1}{1 + 2 x ^{2}} d x

\int_{- 1}^{2} 4 x^{3} + 8 x + 3 d x

\int_{0}^{π} x^{2} \cdot cos (n \cdot x) d x

\int_{0}^{2} x^{2} e^{3 x} d x