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受欢迎的微积分 >

integral of sin(x)cos(x)sqrt(1-sin^4(x))

解答

\int sin (x) cos (x) 1 - sin^{4} (x) d x

解答

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

求解步骤

\int sin (x) cos (x) 1 - sin^{4} (x) d x

使用换元积分法

= \int u 1 - u^{4} d u

化简

u^{4} : (u^{2})^{2}

= \int u 1 - (u^{2})^{2} d u

使用换元积分法

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

提出常数:

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

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

使用三角换元法

= \frac{1}{2} \cdot \int cos^{2} (w) d w

使用三角恒等式改写

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

提出常数:

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

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

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

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

\int 1 d w = w

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

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

代回

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

化简

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

= \frac{1}{4} (arcsin (sin^{2} (x)) + \frac{1}{2} sin (2 arcsin (sin^{2} (x))))

解答补常数

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

作图

流行的例子

(\partial)/(\partial x)((3y)*(2x)^{-1})

\frac{\partial}{\partial x} ((3 y) \cdot (2 x)^{- 1})

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p a r i t y f (x) = sin (tan (2 x))

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\int 3 x y

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\int \frac{1}{( 2 - x ) 1 - x} d x