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受欢迎的微积分 >

integral of 1/(t^2sqrt(16-t^2))

解答

\int \frac{1}{t ^{2} 16 - t ^{2}} d t

解答

- \frac{16 - t ^{2}}{16 t} + C

求解步骤

\int \frac{1}{t ^{2} 16 - t ^{2}} d t

使用三角换元法

= \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

提出常数:

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

= \frac{1}{16} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

化简

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{16} \cdot \int csc^{2} (u) d u

使用积分常见定则:

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

= \frac{1}{16} (- cot (u))

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

代回

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

化简

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

= - \frac{16 - t ^{2}}{16 t}

解答补常数

= - \frac{16 - t ^{2}}{16 t} + C

作图

流行的例子

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s l o p e y = 2 - x^{3}, (1, 1)

y^{^{'}} = (t + y)^{2} - 1

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\int \frac{3 x ^{3} - 4 x ^{2} - 14 x + 8}{x ^{2} - 3 x + 1} d x

derivative f(x)=|x-9|

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