integral 2+2cos^4(t)

Lösung

integral

2 + 2 cos^{4} (t)

\frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t) + C

Schritte zur Lösung

\int 2 + 2 cos^{4} (t) d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 d t + \int 2 cos^{4} (t) d t

\int 2 d t = 2 t

\int 2 cos^{4} (t) d t = 2 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t)))

= 2 t + 2 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t)))

Vereinfache

2 t + 2 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t))) : \frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t)

= \frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t)

Füge eine Konstante zur Lösung hinzu

= \frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t) + C

in t e g r a l (\frac{1 + 1}{x ^{2}}) e^{\frac{x - 1}{x}}

in t e g r a l \frac{4 x}{( x ^{3} - x ^{2} - x + 1 )}

in t e g r a l x \cdot (x - 1)^{3}

in t e g r a l (x - 1)^{4}

in t e g r a l \frac{v}{( v - 1 ) ^{2}}