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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用换元积分法

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

提出常数:

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

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

将

\frac{u ^{2} + 1}{( 1 + u ^{2} - 2 u ) ^{2}}

用部份分式展开:

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

= 2 \cdot \int \frac{1}{( u - 1 ) ^{2}} + \frac{2}{( u - 1 ) ^{3}} + \frac{2}{( u - 1 ) ^{4}} 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

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

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

\int \frac{2}{( u - 1 ) ^{3}} d u = - \frac{1}{( u - 1 ) ^{2}}

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

= 2 (- \frac{1}{u - 1} - \frac{1}{( u - 1 ) ^{2}} - \frac{2}{3 ( u - 1 ) ^{3}})

u = tan (\frac{x}{2})

代回

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

解答补常数

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

作图

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