integral of sec^9(x)tan^3(x)

解答

\int sec^{9} (x) tan^{3} (x) d x

\frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{9} + C

求解步骤

\int sec^{9} (x) tan^{3} (x) d x

使用三角恒等式改写

= \int (- 1 + sec^{2} (x)) tan (x) sec^{9} (x) d x

使用换元积分法

= \int u^{8} (u^{2} - 1) d u

乘开

u^{8} (u^{2} - 1) : u^{10} - u^{8}

= \int u^{10} - u^{8} 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 u^{10} d u - \int u^{8} d u

\int u^{10} d u = \frac{u ^{11}}{11}

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{11}}{11} - \frac{u ^{9}}{9}

u = sec (x)

代回

= \frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{9}

解答补常数

= \frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{9} + C

\int_{0}^{\frac{π}{4}} x sec^{2} (x) d x

d er i v a t i v e f (x) = \frac{- 4}{x}

n = 1 \sum \infty (\frac{2 π}{3})^{n}

\int \frac{1}{x ( x ^{2} + 2 )} d x

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