integral of (sqrt(16-t^2))/(t^2)

解答

\int \frac{16 - t ^{2}}{t ^{2}} d t

- arcsin (\frac{1}{4} t) - \frac{16 - t ^{2}}{t} + C

求解步骤

\int \frac{16 - t ^{2}}{t ^{2}} d t

使用三角换元法

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

使用三角恒等式改写

= \int - 1 + csc^{2} (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

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

\int 1 d u = u

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

= - u - cot (u)

u = arcsin (\frac{1}{4} t)

代回

= - arcsin (\frac{1}{4} t) - cot (arcsin (\frac{1}{4} t))

化简

- arcsin (\frac{1}{4} t) - cot (arcsin (\frac{1}{4} t)) : - arcsin (\frac{1}{4} t) - \frac{16 - t ^{2}}{t}

= - arcsin (\frac{1}{4} t) - \frac{16 - t ^{2}}{t}

解答补常数

= - arcsin (\frac{1}{4} t) - \frac{16 - t ^{2}}{t} + C

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