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受欢迎的微积分 >

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

解答

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

解答

\frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1)) + C

求解步骤

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

使用三角换元法

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

使用换元积分法

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

对

v^{2} + 1 - v

配方:

(v - \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{1}{( v - \frac{1}{2} ) ^{2} + \frac{3}{4}} d v

使用换元积分法

= \int \frac{4}{4 w ^{2} + 3} d w

提出常数:

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

= 4 \cdot \int \frac{1}{4 w ^{2} + 3} d w

使用换元积分法

= 4 \cdot \int \frac{1}{2 3 ( t ^{2} + 1 )} d t

提出常数:

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

= 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{t ^{2} + 1} d t

使用积分常见定则:

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

= 4 \cdot \frac{1}{2 3} arctan (t)

代回

= 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (\frac{arcsin ( x )}{2}) - \frac{1}{2}))

化简

4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (\frac{arcsin ( x )}{2}) - \frac{1}{2})) : \frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1))

= \frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1))

解答补常数

= \frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1)) + C

作图

流行的例子

integral of 11csc^2(x)

\int 11 csc^{2} (x) d x

integral of (3x)/((2-5x^2)^2)

\int \frac{3 x}{( 2 - 5 x ^{2} ) ^{2}} d x

integral of 1/x*sin(x)

\int \frac{1}{x} \cdot sin (x) d x

integral of 3sin(5x)cos(4x)

\int 3 sin (5 x) cos (4 x) d x

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

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