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受欢迎的微积分 >

integral of 1/2 sin((2pit)/5)

解答

\int \frac{1}{2} sin (\frac{2 π t}{5}) d t

解答

- \frac{5}{4 π} cos (\frac{2 π t}{5}) + C

求解步骤

\int \frac{1}{2} sin (\frac{2 π t}{5}) 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 sin (\frac{2 π t}{5}) d t

使用换元积分法

= \frac{1}{2} \cdot \int sin (\frac{u}{5}) \frac{1}{2 π} d u

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot \int sin (\frac{u}{5}) d u

使用换元积分法

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot \int sin (v) \cdot 5 d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 \cdot \int sin (v) d v

使用积分常见定则:

\int sin (v) d v = - cos (v)

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 (- cos (v))

代回

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 (- cos (\frac{2 π t}{5}))

化简

\frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 (- cos (\frac{2 π t}{5})) : - \frac{5}{4 π} cos (\frac{2 π t}{5})

= - \frac{5}{4 π} cos (\frac{2 π t}{5})

解答补常数

= - \frac{5}{4 π} cos (\frac{2 π t}{5}) + C

作图

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\int (1 + x)^{- 2} d x

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\int \frac{8}{t ^{2} ( t ^{2} - 4 )} d t

limit as x approaches 2 of 5x^2-4x+3

x \to 2 lim (5 x^{2} - 4 x + 3)

tangent y=9x-4x^2,(-2,-34)

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integral of (cos(pi/(x^5)))/(x^6)

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