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受欢迎的微积分 >

integral of 7/(sqrt(16-4t^2))

解答

\int \frac{7}{16 - 4 t ^{2}} d t

解答

\frac{7}{2} arcsin (\frac{t}{2}) + C

求解步骤

\int \frac{7}{16 - 4 t ^{2}} d t

提出常数:

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

= 7 \cdot \int \frac{1}{16 - 4 t ^{2}} d t

使用三角换元法

= 7 \cdot \int \frac{1}{2} d u

提出常数:

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

= 7 \cdot \frac{1}{2} \cdot \int 1 d u

常数积分:

\int a d x = a x

= 7 \cdot \frac{1}{2} \cdot 1 \cdot u

u = arcsin (\frac{2}{4} t)

代回

= 7 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{4} t)

化简

7 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{4} t) : \frac{7}{2} arcsin (\frac{t}{2})

= \frac{7}{2} arcsin (\frac{t}{2})

解答补常数

= \frac{7}{2} arcsin (\frac{t}{2}) + C

作图

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