integral of (sin(h)7x)/(1+cos(h^2)7x)

Lösung

\int \frac{sin ( h ) 7 x}{1 + cos ( h ^{2} ) 7 x} d x

\frac{sec ^{2} ( h ^{2} )}{7} sin (h) (1 + 7 x cos (h^{2}) - ln 1 + 7 x cos (h^{2})) + C

Schritte zur Lösung

\int \frac{sin ( h ) \cdot 7 x}{1 + cos ( h ^{2} ) \cdot 7 x} d x

Entferne die Konstante:

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

= sin (h) \cdot 7 \cdot \int \frac{x}{1 + cos ( h ^{2} ) \cdot 7 x} d x

Wende U-Substitution an

= sin (h) \cdot 7 \cdot \int \frac{u - 1}{49 u cos ^{2} ( h ^{2} )} d u

Entferne die Konstante:

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

= sin (h) \cdot 7 \cdot \frac{1}{49 cos ^{2} ( h ^{2} )} \cdot \int \frac{u - 1}{u} d u

Multipliziere aus

\frac{u - 1}{u} : 1 - \frac{1}{u}

= sin (h) \cdot 7 \cdot \frac{1}{49 cos ^{2} ( h ^{2} )} \cdot \int 1 - \frac{1}{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

= sin (h) \cdot 7 \cdot \frac{1}{49 cos ^{2} ( h ^{2} )} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= sin (h) \cdot 7 \cdot \frac{1}{49 cos ^{2} ( h ^{2} )} (u - ln ∣ u ∣)

Setze in

u = 1 + cos (h^{2}) \cdot 7 x

ein

= sin (h) \cdot 7 \cdot \frac{1}{49 cos ^{2} ( h ^{2} )} (1 + cos (h^{2}) \cdot 7 x - ln 1 + cos (h^{2}) \cdot 7 x)

Vereinfache

sin (h) \cdot 7 \cdot \frac{1}{49 cos ^{2} ( h ^{2} )} (1 + cos (h^{2}) \cdot 7 x - ln 1 + cos (h^{2}) \cdot 7 x) : \frac{sec ^{2} ( h ^{2} )}{7} sin (h) (1 + 7 x cos (h^{2}) - ln 1 + 7 x cos (h^{2}))

= \frac{sec ^{2} ( h ^{2} )}{7} sin (h) (1 + 7 x cos (h^{2}) - ln 1 + 7 x cos (h^{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{sec ^{2} ( h ^{2} )}{7} sin (h) (1 + 7 x cos (h^{2}) - ln 1 + 7 x cos (h^{2})) + C

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