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受欢迎的微积分 >

integral of (cos(x))/(5sin^2(x)+6sin(x))

解答

\int \frac{cos ( x )}{5 sin ^{2} ( x ) + 6 sin ( x )} d x

解答

- \frac{1}{6} (ln \frac{5}{3} sin (x) + 2 - ln (\frac{5}{3} ∣ sin (x) ∣)) + C

求解步骤

\int \frac{cos ( x )}{5 sin ^{2} ( x ) + 6 sin ( x )} d x

使用换元积分法

= \int \frac{1}{5 u ^{2} + 6 u} d u

对

5 u^{2} + 6 u

配方:

5 (u + \frac{3}{5})^{2} - \frac{9}{5}

= \int \frac{1}{5 ( u + \frac{3}{5} ) ^{2} - \frac{9}{5}} d u

使用换元积分法

= \int \frac{5}{25 v ^{2} - 9} d v

提出常数:

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

= 5 \cdot \int \frac{1}{25 v ^{2} - 9} d v

使用换元积分法

= 5 \cdot \int \frac{1}{15 ( w ^{2} - 1 )} d w

提出常数:

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

= 5 \cdot \frac{1}{15} \cdot \int \frac{1}{w ^{2} - 1} d w

因式分解

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

= 5 \cdot \frac{1}{15} \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

= 5 \cdot \frac{1}{15} (- \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}

= 5 \cdot \frac{1}{15} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代回

= 5 \cdot \frac{1}{15} (- (\frac{ln \frac{5}{3} ( sin ( x ) + \frac{3}{5} ) + 1}{2} - \frac{ln \frac{5}{3} ( sin ( x ) + \frac{3}{5} ) - 1}{2}))

化简

5 \cdot \frac{1}{15} (- (\frac{ln \frac{5}{3} ( sin ( x ) + \frac{3}{5} ) + 1}{2} - \frac{ln \frac{5}{3} ( sin ( x ) + \frac{3}{5} ) - 1}{2})) : - \frac{1}{6} (ln \frac{5}{3} sin (x) + 2 - ln (\frac{5}{3} ∣ sin (x) ∣))

= - \frac{1}{6} (ln \frac{5}{3} sin (x) + 2 - ln (\frac{5}{3} ∣ sin (x) ∣))

解答补常数

= - \frac{1}{6} (ln \frac{5}{3} sin (x) + 2 - ln (\frac{5}{3} ∣ sin (x) ∣)) + C

作图

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