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受欢迎的微积分 >

integral of (x^2)/(sqrt((4-9x^2)))

解答

\int \frac{x ^{2}}{( 4 - 9 x ^{2} )} d x

解答

\frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x))) + C

求解步骤

\int \frac{x ^{2}}{4 - 9 x ^{2}} d x

使用三角换元法

= \int \frac{4 sin ^{2} ( u )}{27} d u

提出常数:

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

= \frac{4}{27} \cdot \int sin^{2} (u) d u

使用三角恒等式改写

= \frac{4}{27} \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{4}{27} \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{4}{27} \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{4}{27} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

u = arcsin (\frac{3}{2} x)

代回

= \frac{4}{27} \cdot \frac{1}{2} (arcsin (\frac{3}{2} x) - \frac{1}{2} sin (2 arcsin (\frac{3}{2} x)))

化简

\frac{4}{27} \cdot \frac{1}{2} (arcsin (\frac{3}{2} x) - \frac{1}{2} sin (2 arcsin (\frac{3}{2} x))) : \frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x)))

= \frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x)))

解答补常数

= \frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x))) + C

作图

流行的例子

integral of x^3sqrt(16-4x^2)

\int x^{3} 16 - 4 x^{2} d x

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

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

integral of (5x^4-2x^5)

\int (5 x^{4} - 2 x^{5}) d x

integral of (1/((5x-3)(3-4x)))

\int (\frac{1}{( 5 x - 3 ) ( 3 - 4 x )}) d x

integral of (arctan(2t))/(1+4t^2)

\int \frac{arctan ( 2 t )}{1 + 4 t ^{2}} d t