integral of (1-sin(x))^{1/2}

Lösung

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

2 (- cos (\frac{x}{2}) - sin (\frac{x}{2})) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{2 ( u - 1 )}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{u - 1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

Multipliziere aus

\frac{u - 1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} : \frac{u}{( 1 + u ^{2} ) ^{\frac{3}{2}}} - \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}}

= 2 \cdot \int \frac{u}{( 1 + u ^{2} ) ^{\frac{3}{2}}} - \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} 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

= 2 (\int \frac{u}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u - \int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u)

\int \frac{u}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u = - \frac{1}{( 1 + u ^{2} ) ^{\frac{1}{2}}}

\int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u = \frac{u}{( 1 + u ^{2} ) ^{\frac{1}{2}}}

= 2 (- \frac{1}{( 1 + u ^{2} ) ^{\frac{1}{2}}} - \frac{u}{( 1 + u ^{2} ) ^{\frac{1}{2}}})

Setze in

u = tan (\frac{x}{2})

ein

= 2 - \frac{1}{( 1 + tan ^{2} ( \frac{x}{2} ) ) ^{\frac{1}{2}}} - \frac{tan ( \frac{x}{2} )}{( 1 + tan ^{2} ( \frac{x}{2} ) ) ^{\frac{1}{2}}}

Vereinfache

2 - \frac{1}{( 1 + tan ^{2} ( \frac{x}{2} ) ) ^{\frac{1}{2}}} - \frac{tan ( \frac{x}{2} )}{( 1 + tan ^{2} ( \frac{x}{2} ) ) ^{\frac{1}{2}}} : 2 (- cos (\frac{x}{2}) - sin (\frac{x}{2}))

= 2 (- cos (\frac{x}{2}) - sin (\frac{x}{2}))

Füge eine Konstante zur Lösung hinzu

= 2 (- cos (\frac{x}{2}) - sin (\frac{x}{2})) + C

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}

t an g e n t f (x) = - x^{2} + 16 x + 63, at x = 2

\int ln (4 x) d x

\frac{d}{d x} (\frac{1}{1 - cos ( x )})

\int \frac{2 x}{x ^{2} - 4 x + 13} d x