integral of 1/(1-sin(5x))

解

\int \frac{1}{1 - sin ( 5 x )} d x

\frac{1}{5} tan (5 x) + \frac{1}{5} sec (5 x) + C

解答ステップ

\int \frac{1}{1 - sin ( 5 x )} d x

1 - sin (5 x)

の共役で乗じる

\frac{cos ^{2} ( 5 x )}{1 + sin ( 5 x )}

= \int \frac{1}{\frac{c o s ^{2} ( 5 x )}{1 + s i n ( 5 x )}} d x

拡張

\frac{1}{\frac{c o s ^{2} ( 5 x )}{1 + s i n ( 5 x )}} : \frac{1}{cos ^{2} ( 5 x )} + \frac{sin ( 5 x )}{cos ^{2} ( 5 x )}

= \int \frac{1}{cos ^{2} ( 5 x )} + \frac{sin ( 5 x )}{cos ^{2} ( 5 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{1}{cos ^{2} ( 5 x )} d x + \int \frac{sin ( 5 x )}{cos ^{2} ( 5 x )} d x

\int \frac{1}{cos ^{2} ( 5 x )} d x = \frac{1}{5} tan (5 x)

\int \frac{sin ( 5 x )}{cos ^{2} ( 5 x )} d x = \frac{1}{5} sec (5 x)

= \frac{1}{5} tan (5 x) + \frac{1}{5} sec (5 x)

定数を解答に追加する

= \frac{1}{5} tan (5 x) + \frac{1}{5} sec (5 x) + C