integral of (4sin^3(x)-1)/(sin^2(x))

解

\int \frac{4 sin ^{3} ( x ) - 1}{sin ^{2} ( x )} d x

- 4 cos (x) + cot (x) + C

解答ステップ

\int \frac{4 sin ^{3} ( x ) - 1}{sin ^{2} ( x )} d x

拡張

\frac{4 sin ^{3} ( x ) - 1}{sin ^{2} ( x )} : 4 sin (x) - \frac{1}{sin ^{2} ( x )}

= \int 4 sin (x) - \frac{1}{sin ^{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 4 sin (x) d x - \int \frac{1}{sin ^{2} ( x )} d x

\int 4 sin (x) d x = - 4 cos (x)

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

= - 4 cos (x) - (- cot (x))

簡素化

= - 4 cos (x) + cot (x)

定数を解答に追加する

= - 4 cos (x) + cot (x) + C