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受欢迎的微积分 >

integral of e^xsqrt(64-e^{2x)}

解答

\int e^{x} 64 - e^{2 x} d x

解答

32 (arcsin (\frac{1}{8} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} e^{x}))) + C

求解步骤

\int e^{x} 64 - e^{2 x} d x

使用换元积分法

= \int 64 - u^{2} d u

使用三角换元法

= \int 64 cos^{2} (v) d v

提出常数:

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

= 64 \cdot \int cos^{2} (v) d v

使用三角恒等式改写

= 64 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

提出常数:

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

= 64 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 64 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 64 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= 64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} e^{x})))

化简

64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} e^{x}))) : 32 (arcsin (\frac{1}{8} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} e^{x})))

= 32 (arcsin (\frac{1}{8} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} e^{x})))

解答补常数

= 32 (arcsin (\frac{1}{8} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} e^{x}))) + C

作图

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