integral of csc^3(x)cot^2(x)

解

\int csc^{3} (x) cot^{2} (x) d x

- \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x) + C

解答ステップ

\int csc^{3} (x) cot^{2} (x) d x

三角関数の公式を使用して書き換える

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

拡張

\frac{1 - sin ^{2} ( x )}{sin ^{5} ( x )} : \frac{1}{sin ^{5} ( x )} - \frac{1}{sin ^{3} ( x )}

= \int \frac{1}{sin ^{5} ( x )} - \frac{1}{sin ^{3} ( 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}{sin ^{5} ( x )} d x - \int \frac{1}{sin ^{3} ( x )} d x

\int \frac{1}{sin ^{5} ( x )} d x = - \frac{cos ( x ) csc ^{4} ( x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (x) csc (x) + \frac{1}{2} ln tan (\frac{x}{2}))

\int \frac{1}{sin ^{3} ( x )} d x = \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

= - \frac{cos ( x ) csc ^{4} ( x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (x) csc (x) + \frac{1}{2} ln tan (\frac{x}{2})) - \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

簡素化

- \frac{cos ( x ) csc ^{4} ( x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (x) csc (x) + \frac{1}{2} ln tan (\frac{x}{2})) - \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) : - \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x)

= - \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x)

定数を解答に追加する

= - \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x) + C