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

解答

\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}

求解步骤

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

使用三角恒等式改写

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

使用三角恒等式改写

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

使用换元积分法

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

使用积分加法定则:

\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}

化简

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

\int_{0}^{2} 5 x^{2} - 8 x + 4 d x

\int_{- 1}^{1} 1 + x + x^{2} + x^{3} d x

\int_{- 1}^{1} (1 - x) 1 - x^{2} d x

\int_{1}^{4} 1 + x d x

\int_{0}^{6} 6 y^{3} - 6 y^{2} d y