integral of sin^4(θ)

Lösung

\int sin^{4} (θ) d θ

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

Schritte zur Lösung

\int sin^{4} (θ) d θ

Wende integrale Reduktion an

= - \frac{cos ( θ ) sin ^{3} ( θ )}{4} + \frac{3}{4} \cdot \int sin^{2} (θ) d θ

\int sin^{2} (θ) d θ = \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

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

Vereinfache

- \frac{cos ( θ ) sin ^{3} ( θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : - \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ))

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t y = x^{5} e^{- x}, (5, \frac{3125}{e ^{5}})

y^{^{''}} + 6 y^{^{'}} + 8 y = 6 t e^{3 t}

\int \frac{2 a y}{x ^{2} + y ^{2}} d x

x \to 1 lim (x^{2} + 2 x + 3)

d er i v a t i v e cos (\frac{x}{x + 2})