integral of (sin(x))/(1-sin(x))

解

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

sec (x) + tan (x) - x + C

解答ステップ

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

1 - sin (x)

の共役で乗じる

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

= \int \frac{sin ( x )}{\frac{c o s ^{2} ( x )}{1 + s i n ( x )}} d x

拡張

\frac{sin ( x )}{\frac{c o s ^{2} ( x )}{1 + s i n ( x )}} : \frac{sin ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \int \frac{sin ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x

総和規則を適用する:

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

= \int \frac{sin ( x )}{cos ^{2} ( x )} d x + \int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x

\int \frac{sin ( x )}{cos ^{2} ( x )} d x = sec (x)

\int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x = tan (x) - x

= sec (x) + tan (x) - x

定数を解答に追加する

= sec (x) + tan (x) - x + C