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受欢迎的微积分 >

integral of 1/(-4cos(x)+sin(x)-3)

解答

\int \frac{1}{- 4 cos ( x ) + sin ( x ) - 3} d x

解答

- \frac{1}{2 2} (ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} + 1 - ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} - 1) + C

求解步骤

\int \frac{1}{- 4 cos ( x ) + sin ( x ) - 3} d x

使用换元积分法

= \int \frac{2}{u ^{2} + 2 u - 7} d u

提出常数:

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

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

因式分解

u^{2} + 2 u - 7 : - (- u^{2} - 2 u + 7)

= 2 \cdot \int \frac{1}{- ( - u ^{2} - 2 u + 7 )} d u

提出常数:

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

= 2 (- \int \frac{1}{- u ^{2} - 2 u + 7} d u)

对

- u^{2} - 2 u + 7

配方:

- (u + 1)^{2} + 8

= 2 (- \int \frac{1}{- ( u + 1 ) ^{2} + 8} d u)

使用换元积分法

= 2 (- \int \frac{1}{- v ^{2} + 8} d v)

使用换元积分法

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

提出常数:

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

= 2 (- \frac{1}{2 2} \cdot \int \frac{1}{- w ^{2} + 1} d w)

使用积分常见定则:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 (- \frac{1}{2 2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代回

= 2 - \frac{1}{2 2} \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2 2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2 2} - 1}{2}

化简

2 - \frac{1}{2 2} \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2 2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2 2} - 1}{2} : - \frac{1}{2 2} (ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} + 1 - ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} - 1)

= - \frac{1}{2 2} (ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} + 1 - ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} - 1)

解答补常数

= - \frac{1}{2 2} (ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} + 1 - ln \frac{tan ( \frac{x}{2} ) + 1}{2 2} - 1) + C

作图

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