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受欢迎的微积分 >

integral of 1/((sin(x))^3)

解答

\int \frac{1}{( sin ( x ) ) ^{3}} d x

解答

\frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) + C

求解步骤

\int \frac{1}{( sin ( x ) ) ^{3}} d x

使用三角恒等式改写

= \int csc^{3} (x) d x

使用换元积分法

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

提出常数:

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

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

乘开

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= \frac{1}{4} (- \frac{1}{2 u ^{2}} + 2 ln ∣ u ∣ + \frac{u ^{2}}{2})

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

代回

= \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + 2 ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2})

化简

\frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + 2 ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2}) : \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

= \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

解答补常数

= \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) + C

作图

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