integral from-2x to x of 1/(csc^5(t))

解

\int_{- 2 x}^{x} \frac{1}{csc ^{5} ( t )} d t

- cos (x) + cos (2 x) + \frac{2 ( cos ^{3} ( x ) - cos ^{3} ( 2 x ) )}{3} - \frac{cos ^{5} ( x ) - cos ^{5} ( 2 x )}{5}

解答ステップ

\int_{- 2 x}^{x} \frac{1}{csc ^{5} ( t )} d t

三角関数の公式を使用して書き換える

= \int_{- 2 x}^{x} sin^{5} (t) d t

三角関数の公式を使用して書き換える

= \int_{- 2 x}^{x} (1 - cos^{2} (t))^{2} sin (t) d t

u置換積分法を適用する

= \int_{c o s (- 2 x)}^{c o s (x)} - 1 + 2 u^{2} - u^{4} d u

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_{c o s (- 2 x)}^{c o s (x)} 1 d u + \int_{c o s (- 2 x)}^{c o s (x)} 2 u^{2} d u - \int_{c o s (- 2 x)}^{c o s (x)} u^{4} d u

\int_{c o s (- 2 x)}^{c o s (x)} 1 d u = cos (x) - cos (2 x)

\int_{c o s (- 2 x)}^{c o s (x)} 2 u^{2} d u = \frac{2 ( cos ^{3} ( x ) - cos ^{3} ( 2 x ) )}{3}

\int_{c o s (- 2 x)}^{c o s (x)} u^{4} d u = \frac{cos ^{5} ( x ) - cos ^{5} ( 2 x )}{5}

= - (cos (x) - cos (2 x)) + \frac{2 ( cos ^{3} ( x ) - cos ^{3} ( 2 x ) )}{3} - \frac{cos ^{5} ( x ) - cos ^{5} ( 2 x )}{5}

簡素化

= - cos (x) + cos (2 x) + \frac{2 ( cos ^{3} ( x ) - cos ^{3} ( 2 x ) )}{3} - \frac{cos ^{5} ( x ) - cos ^{5} ( 2 x )}{5}