integral from sin(x) to 1 of sqrt(1+x^2)

Lösung

\int_{s i n (x)}^{1} 1 + x^{2} d x

\frac{2 - sin ( x ) sin ^{2} ( x ) + 1 + ln ( 1 + 2 ) - ln sin ( x ) + sin ^{2} ( x ) + 1}{2}

Schritte zur Lösung

\int_{s i n (x)}^{1} 1 + x^{2} d x

Trigonometrische Substitution anwenden

= \int_{a r c t a n (s i n (x))}^{\frac{π}{4}} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (s i n (x))}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{a r c t a n (s i n (x))}^{\frac{π}{4}} sec (u) d u

\int_{a r c t a n (s i n (x))}^{\frac{π}{4}} sec (u) d u = ln (1 + 2) - ln sin (x) + 1 + sin^{2} (x)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (s i n (x))}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln sin (x) + 1 + sin^{2} (x))

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (s i n (x))}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln sin (x) + 1 + sin^{2} (x)) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (s i n (x))}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln sin (x) + 1 + sin^{2} (x))

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (s i n (x))}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln sin (x) + 1 + sin^{2} (x))

Berechne die Grenzen:

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

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

Vereinfache

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

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