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受欢迎的微积分 >

integral of (cos(x))/(sin^2(x)+sin(x)-6)

解答

\int \frac{cos ( x )}{sin ^{2} ( x ) + sin ( x ) - 6} d x

解答

- \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} sin (x) - ln - \frac{4}{5} + \frac{2}{5} sin (x)) + C

求解步骤

\int \frac{cos ( x )}{sin ^{2} ( x ) + sin ( x ) - 6} d x

使用换元积分法

= \int \frac{1}{u ^{2} + u - 6} d u

对

u^{2} + u - 6

配方:

(u + \frac{1}{2})^{2} - \frac{25}{4}

= \int \frac{1}{( u + \frac{1}{2} ) ^{2} - \frac{25}{4}} d u

使用换元积分法

= \int \frac{4}{4 v ^{2} - 25} d v

提出常数:

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

= 4 \cdot \int \frac{1}{4 v ^{2} - 25} d v

使用换元积分法

= 4 \cdot \int \frac{1}{10 ( w ^{2} - 1 )} d w

提出常数:

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

= 4 \cdot \frac{1}{10} \cdot \int \frac{1}{w ^{2} - 1} d w

因式分解

w^{2} - 1 : - (- w^{2} + 1)

= 4 \cdot \frac{1}{10} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

提出常数:

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

= 4 \cdot \frac{1}{10} (- \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}

= 4 \cdot \frac{1}{10} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代回

= 4 \cdot \frac{1}{10} (- (\frac{ln \frac{2}{5} ( sin ( x ) + \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( sin ( x ) + \frac{1}{2} ) - 1}{2}))

化简

4 \cdot \frac{1}{10} (- (\frac{ln \frac{2}{5} ( sin ( x ) + \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( sin ( x ) + \frac{1}{2} ) - 1}{2})) : - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} sin (x) - ln - \frac{4}{5} + \frac{2}{5} sin (x))

= - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} sin (x) - ln - \frac{4}{5} + \frac{2}{5} sin (x))

解答补常数

= - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} sin (x) - ln - \frac{4}{5} + \frac{2}{5} sin (x)) + C

作图

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