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受欢迎的微积分 >

integral of (sqrt(9-t^2))/(2t^2)

解答

\int \frac{9 - t ^{2}}{2 t ^{2}} d t

解答

\frac{1}{2} (- arcsin (\frac{1}{3} t) - \frac{9 - t ^{2}}{t}) + C

求解步骤

\int \frac{9 - t ^{2}}{2 t ^{2}} d t

提出常数:

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

= \frac{1}{2} \cdot \int \frac{9 - t ^{2}}{t ^{2}} d t

使用三角换元法

= \frac{1}{2} \cdot \int cot^{2} (u) d u

使用三角恒等式改写

= \frac{1}{2} \cdot \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

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

\int 1 d u = u

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

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

u = arcsin (\frac{1}{3} t)

代回

= \frac{1}{2} (- arcsin (\frac{1}{3} t) - cot (arcsin (\frac{1}{3} t)))

化简

\frac{1}{2} (- arcsin (\frac{1}{3} t) - cot (arcsin (\frac{1}{3} t))) : \frac{1}{2} (- arcsin (\frac{1}{3} t) - \frac{9 - t ^{2}}{t})

= \frac{1}{2} (- arcsin (\frac{1}{3} t) - \frac{9 - t ^{2}}{t})

解答补常数

= \frac{1}{2} (- arcsin (\frac{1}{3} t) - \frac{9 - t ^{2}}{t}) + C

作图

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