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受欢迎的微积分 >

integral of 11arcsin(sqrt(2x))

解答

\int 11 arcsin (2 x) d x

解答

11 (x arcsin (2 x) - \frac{1}{4} (arcsin (2 x) - \frac{1}{2} sin (2 arcsin (2 x)))) + C

求解步骤

\int 11 arcsin (2 x) d x

提出常数:

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

= 11 \cdot \int arcsin (2 x) d x

使用换元积分法

= 11 \cdot \int arcsin (u) \frac{1}{2} d u

提出常数:

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

= 11 \cdot \frac{1}{2} \cdot \int arcsin (u) d u

使用换元积分法

= 11 \cdot \frac{1}{2} \cdot \int arcsin (v) \cdot 2 v d v

提出常数:

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

= 11 \cdot \frac{1}{2} \cdot 2 \cdot \int arcsin (v) v d v

使用分布积分法

= 11 \cdot \frac{1}{2} \cdot 2 (\frac{1}{2} v^{2} arcsin (v) - \int \frac{v ^{2}}{2 1 - v ^{2}} d v)

\int \frac{v ^{2}}{2 1 - v ^{2}} d v = \frac{1}{4} (arcsin (v) - \frac{1}{2} sin (2 arcsin (v)))

= 11 \cdot \frac{1}{2} \cdot 2 (\frac{1}{2} v^{2} arcsin (v) - \frac{1}{4} (arcsin (v) - \frac{1}{2} sin (2 arcsin (v))))

代回

= 11 \cdot \frac{1}{2} \cdot 2 (\frac{1}{2} (2 x)^{2} arcsin (2 x) - \frac{1}{4} (arcsin (2 x) - \frac{1}{2} sin (2 arcsin (2 x))))

化简

11 \cdot \frac{1}{2} \cdot 2 (\frac{1}{2} (2 x)^{2} arcsin (2 x) - \frac{1}{4} (arcsin (2 x) - \frac{1}{2} sin (2 arcsin (2 x)))) : 11 (x arcsin (2 x) - \frac{1}{4} (arcsin (2 x) - \frac{1}{2} sin (2 arcsin (2 x))))

= 11 (x arcsin (2 x) - \frac{1}{4} (arcsin (2 x) - \frac{1}{2} sin (2 arcsin (2 x))))

解答补常数

= 11 (x arcsin (2 x) - \frac{1}{4} (arcsin (2 x) - \frac{1}{2} sin (2 arcsin (2 x)))) + C

作图

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