integral from 0 to x of sqrt(1-sin(x))

Lösung

\int_{0}^{x} 1 - sin (x) d x

2 sin (x) + 1 - 2

Schritte zur Lösung

\int_{0}^{x} 1 - sin (x) d x

Wende U-Substitution an

= \int_{0}^{s i n (x)} \frac{1}{u + 1} d u

Wende U-Substitution an

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

Vereinfache

\frac{1}{v} : \frac{1}{v}

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

Vereinfache

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

Wende die Potenzregel an

= [2 v^{\frac{1}{2}}]_{1}^{s i n (x) + 1}

Wende Radikal Regel an:

a^{\frac{1}{n}} = n a

= [2 v]_{1}^{s i n (x) + 1}

Berechne die Grenzen:

2 sin (x) + 1 - 2

= 2 sin (x) + 1 - 2

\int_{0}^{\frac{1}{2}} 3 x d x

\int_{\frac{1}{2}}^{2} (5 - \frac{5}{x ^{2}}) d x

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\int_{1}^{\infty} x \cdot \frac{3}{x ^{4}} d x