integral from 0 to pi/(12) of (sin(6x))/(1+cos^2(6x))

Lösung

\int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{1 + cos ^{2} ( 6 x )} d x

\frac{π}{24}

Dezimale

0.13089 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{1 + cos ^{2} ( 6 x )} d x

Wende U-Substitution an

= \int_{0}^{\frac{π}{2}} \frac{sin ( u )}{6 ( 1 + cos ^{2} ( u ) )} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} \frac{sin ( u )}{1 + cos ^{2} ( u )} d u

Wende U-Substitution an

= \frac{1}{6} \cdot \int_{1}^{0} - \frac{1}{1 + v ^{2}} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{6} (- \int_{0}^{1} - \frac{1}{1 + v ^{2}} d v)

Entferne die Konstante:

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

= \frac{1}{6} (- (- \int_{0}^{1} \frac{1}{1 + v ^{2}} d v))

Nutze das gemeinsame Integral :

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= \frac{1}{6} (- (- [arctan (v)]_{0}^{1}))

Vereinfache

= \frac{1}{6} [arctan (v)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{6} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{24}

3 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

\frac{d}{d x} (2 sin (x^{2} - 1))

\int \frac{1}{- x ^{2} + 4 x + 12} d x

n = 1 \sum \infty \frac{8}{n ^{5}}

t an g e n t e^{- 2 x}