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受欢迎的微积分 >

integral of 1/((3+sin(x)))

解答

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

解答

\frac{1}{2} arctan (\frac{1}{2 2} (3 tan (\frac{x}{2}) + 1)) + C

求解步骤

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

使用换元积分法

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

对

3 (1 + u^{2}) + 2 u

配方:

3 (u + \frac{1}{3})^{2} + \frac{8}{3}

= 2 \cdot \int \frac{1}{3 ( u + \frac{1}{3} ) ^{2} + \frac{8}{3}} d u

使用换元积分法

= 2 \cdot \int \frac{3}{9 v ^{2} + 8} d v

提出常数:

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

= 2 \cdot 3 \cdot \int \frac{1}{9 v ^{2} + 8} d v

使用换元积分法

= 2 \cdot 3 \cdot \int \frac{1}{6 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 \cdot 3 \cdot \frac{1}{6 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 = arctan (w)

= 2 \cdot 3 \cdot \frac{1}{6 2} arctan (w)

代回

= 2 \cdot 3 \cdot \frac{1}{6 2} arctan (\frac{3}{2 2} (tan (\frac{x}{2}) + \frac{1}{3}))

化简

2 \cdot 3 \cdot \frac{1}{6 2} arctan (\frac{3}{2 2} (tan (\frac{x}{2}) + \frac{1}{3})) : \frac{1}{2} arctan (\frac{1}{2 2} (3 tan (\frac{x}{2}) + 1))

= \frac{1}{2} arctan (\frac{1}{2 2} (3 tan (\frac{x}{2}) + 1))

解答补常数

= \frac{1}{2} arctan (\frac{1}{2 2} (3 tan (\frac{x}{2}) + 1)) + C

作图

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