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受欢迎的微积分 >

integral of xsec^4(x^2+1)tan^7(x^2+1)

解答

\int x sec^{4} (x^{2} + 1) tan^{7} (x^{2} + 1) d x

解答

\frac{1}{2} (\frac{tan ^{8} ( x ^{2} + 1 )}{8} + \frac{tan ^{10} ( x ^{2} + 1 )}{10}) + C

求解步骤

\int x sec^{4} (x^{2} + 1) tan^{7} (x^{2} + 1) d x

使用换元积分法

= \int \frac{tan ^{7} ( u ) sec ^{4} ( u )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int tan^{7} (u) sec^{4} (u) d u

使用三角恒等式改写

= \frac{1}{2} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{7} (u) d u

使用换元积分法

= \frac{1}{2} \cdot \int v^{7} (1 + v^{2}) d v

乘开

v^{7} (1 + v^{2}) : v^{7} + v^{9}

= \frac{1}{2} \cdot \int v^{7} + v^{9} 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}{2} (\int v^{7} d v + \int v^{9} d v)

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

\int v^{9} d v = \frac{v ^{10}}{10}

= \frac{1}{2} (\frac{v ^{8}}{8} + \frac{v ^{10}}{10})

代回

= \frac{1}{2} (\frac{tan ^{8} ( x ^{2} + 1 )}{8} + \frac{tan ^{10} ( x ^{2} + 1 )}{10})

解答补常数

= \frac{1}{2} (\frac{tan ^{8} ( x ^{2} + 1 )}{8} + \frac{tan ^{10} ( x ^{2} + 1 )}{10}) + C

作图

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