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受欢迎的微积分 >

integral of tan^2(5x)sec^4(5x)

解答

\int tan^{2} (5 x) sec^{4} (5 x) d x

解答

\frac{1}{5} (\frac{tan ^{3} ( 5 x )}{3} + \frac{tan ^{5} ( 5 x )}{5}) + C

求解步骤

\int tan^{2} (5 x) sec^{4} (5 x) d x

使用换元积分法

= \int tan^{2} (u) sec^{4} (u) \frac{1}{5} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int tan^{2} (u) sec^{4} (u) d u

使用三角恒等式改写

= \frac{1}{5} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

使用换元积分法

= \frac{1}{5} \cdot \int v^{2} (1 + v^{2}) d v

乘开

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{1}{5} \cdot \int v^{2} + v^{4} d v

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

= \frac{1}{5} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{5} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

代回

= \frac{1}{5} (\frac{tan ^{3} ( 5 x )}{3} + \frac{tan ^{5} ( 5 x )}{5})

解答补常数

= \frac{1}{5} (\frac{tan ^{3} ( 5 x )}{3} + \frac{tan ^{5} ( 5 x )}{5}) + C

作图

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