integral of 1/(a^2sin^2(x)+b^2cos^2(x))

解答

\int \frac{1}{a ^{2} sin ^{2} ( x ) + b ^{2} cos ^{2} ( x )} d x

\frac{1}{ab} arctan (\frac{a}{b} tan (x)) + C

求解步骤

\int \frac{1}{a ^{2} sin ^{2} ( x ) + b ^{2} cos ^{2} ( x )} d x

使用三角换元法

= \int \frac{1}{a ^{2} u ^{2} + b ^{2}} d u

使用换元积分法

= \int \frac{1}{ab ( v ^{2} + 1 )} d v

提出常数:

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

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

使用积分常见定则:

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

= \frac{1}{ab} arctan (v)

代回

= \frac{1}{ab} arctan (\frac{a}{b} tan (x))

解答补常数

= \frac{1}{ab} arctan (\frac{a}{b} tan (x)) + C

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + a ^{2}})

d er i v a t i v e \frac{x ^{2}}{3 ( 1 - x )}

d er i v a t i v e (4 x - 3)^{4}

\int x arctan (7 x) d x

z \to 0 lim (\frac{sin ( z )}{z})