integral of cot^2(6x^2)x

解答

\int cot^{2} (6 x^{2}) x d x

\frac{1}{2} (- x^{2} - \frac{1}{6} cot (6 x^{2})) + C

求解步骤

\int cot^{2} (6 x^{2}) x d x

使用换元积分法

= \int \frac{cot ^{2} ( 6 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 cot^{2} (6 u) d u

使用三角恒等式改写

= \frac{1}{2} \cdot \int - 1 + csc^{2} (6 u) 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

= \frac{1}{2} (- \int 1 d u + \int csc^{2} (6 u) d u)

\int 1 d u = u

\int csc^{2} (6 u) d u = - \frac{1}{6} cot (6 u)

= \frac{1}{2} (- u - \frac{1}{6} cot (6 u))

u = x^{2}

代回

= \frac{1}{2} (- x^{2} - \frac{1}{6} cot (6 x^{2}))

解答补常数

= \frac{1}{2} (- x^{2} - \frac{1}{6} cot (6 x^{2})) + C

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