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受欢迎的微积分 >

integral of cot^{-9}(x)

解答

\int cot^{- 9} (x) d x

解答

ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8} + C

求解步骤

\int cot^{- 9} (x) d x

化简

cot^{- 9} (x) : tan^{9} (x)

= \int tan^{9} (x) d x

使用三角恒等式改写

= \int (- 1 + sec^{2} (x))^{4} tan (x) d x

使用换元积分法

= \int \frac{( - 1 + u ^{2} ) ^{4}}{u} d u

乘开

\frac{( - 1 + u ^{2} ) ^{4}}{u} : \frac{1}{u} - 4 u + 6 u^{3} - 4 u^{5} + u^{7}

= \int \frac{1}{u} - 4 u + 6 u^{3} - 4 u^{5} + u^{7} 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 \frac{1}{u} d u - \int 4 u d u + \int 6 u^{3} d u - \int 4 u^{5} d u + \int u^{7} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int 4 u d u = 2 u^{2}

\int 6 u^{3} d u = \frac{3 u ^{4}}{2}

\int 4 u^{5} d u = \frac{2 u ^{6}}{3}

\int u^{7} d u = \frac{u ^{8}}{8}

= ln ∣ u ∣ - 2 u^{2} + \frac{3 u ^{4}}{2} - \frac{2 u ^{6}}{3} + \frac{u ^{8}}{8}

u = sec (x)

代回

= ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8}

解答补常数

= ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8} + C

作图

流行的例子

derivative of 1-2x^2

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

integral of x/((16-x^2)^2)

\int \frac{x}{( 16 - x ^{2} ) ^{2}} d x

tangent y=(-4x)/(x^2+1),(1,-2)

t an g e n t y = \frac{- 4 x}{x ^{2} + 1}, (1, - 2)

integral of (10)/(sqrt(x+10)+\sqrt{x)}

\int \frac{10}{x + 10 + x} d x

integral from 2 to 1 of (12x^{-5})

\int_{2}^{1} (12 x^{- 5}) d x