integral of-9tan^2(x)

解答

\int - 9 tan^{2} (x) d x

- 9 (- x + tan (x)) + C

求解步骤

\int - 9 tan^{2} (x) d x

提出常数:

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

= - 9 \cdot \int tan^{2} (x) d x

使用三角恒等式改写

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

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

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

\int 1 d x = x

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

= - 9 (- x + tan (x))

解答补常数

= - 9 (- x + tan (x)) + C

t an g e n t f (x) = x, at x = 16

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