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受欢迎的微积分 >

integral of (x^2)/(sqrt(-16x^2+1))

解答

\int \frac{x ^{2}}{- 16 x ^{2} + 1} d x

解答

\frac{1}{128} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x))) + C

求解步骤

\int \frac{x ^{2}}{- 16 x ^{2} + 1} d x

使用三角换元法

= \int \frac{sin ^{2} ( u )}{64} d u

提出常数:

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

= \frac{1}{64} \cdot \int sin^{2} (u) d u

使用三角恒等式改写

= \frac{1}{64} \cdot \int \frac{1 - cos ( 2 u )}{2} d u

提出常数:

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

= \frac{1}{64} \cdot \frac{1}{2} \cdot \int 1 - cos (2 u) d u

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}{64} \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

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

= \frac{1}{64} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

u = arcsin (4 x)

代回

= \frac{1}{64} \cdot \frac{1}{2} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x)))

化简

\frac{1}{64} \cdot \frac{1}{2} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x))) : \frac{1}{128} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x)))

= \frac{1}{128} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x)))

解答补常数

= \frac{1}{128} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x))) + C

作图

流行的例子

integral of 7/(sqrt(16-4t^2))

\int \frac{7}{16 - 4 t ^{2}} d t

integral of 4xsqrt(x+1)

\int 4 x x + 1 d x

integral of 8xe^{7x^2}

\int 8 x e^{7 x^{2}} d x

integral of x/(e^2)

\int \frac{x}{e ^{2}} d x

integral of sin^{-2}(2x)

\int sin^{- 2} (2 x) d x