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受欢迎的微积分 >

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

解答

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

解答

- 2 cos^{2} (\frac{x}{2}) (- tan (\frac{x}{2}) + sin^{2} (\frac{x}{2})) + C

求解步骤

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

使用换元积分法

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

将

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

用部份分式展开:

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

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

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

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

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

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

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

代回

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

化简

- 2 (arctan (tan (\frac{x}{2})) + \frac{1}{tan ^{2} ( \frac{x}{2} ) + 1} - arctan (tan (\frac{x}{2})) - \frac{tan ( \frac{x}{2} )}{1 + tan ^{2} ( \frac{x}{2} )} - \frac{1}{( tan ^{2} ( \frac{x}{2} ) + 1 ) ^{2}}) : - 2 cos^{2} (\frac{x}{2}) (- tan (\frac{x}{2}) + sin^{2} (\frac{x}{2}))

= - 2 cos^{2} (\frac{x}{2}) (- tan (\frac{x}{2}) + sin^{2} (\frac{x}{2}))

解答补常数

= - 2 cos^{2} (\frac{x}{2}) (- tan (\frac{x}{2}) + sin^{2} (\frac{x}{2})) + C

作图

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