integral of tan^3(x)sec^4(x)

解答

\int tan^{3} (x) sec^{4} (x) d x

\frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{6} + C

求解步骤

\int tan^{3} (x) sec^{4} (x) d x

使用三角恒等式改写

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{3} (x) d x

使用换元积分法

= \int u^{3} (1 + u^{2}) d u

乘开

u^{3} (1 + u^{2}) : u^{3} + u^{5}

= \int u^{3} + u^{5} 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^{3} d u + \int u^{5} d u

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

\int u^{5} d u = \frac{u ^{6}}{6}

= \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

u = tan (x)

代回

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{6}

解答补常数

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{6} + C

l a pl a ce t r an s f or m t^{\frac{3}{2}} \cdot e^{- 2 t}

\frac{\partial}{\partial s} (\frac{t}{s})

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\int \frac{1}{5} x d x

\int \frac{2}{x ^{3} x ^{4} - 81} d x