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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

\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

以下の部分分数を得る:

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

総和規則を適用する:

\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}})

代用を戻す

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

= - 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}})

簡素化

- 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}))

定数を解答に追加する

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

グラフ

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