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人気のある微分積分 >

integral of (cos^5(5x+1))/(sin^4(5x+1))

解

\int \frac{cos ^{5} ( 5 x + 1 )}{sin ^{4} ( 5 x + 1 )} d x

解

\frac{1}{5} (- \frac{1}{3} csc^{3} (5 x + 1) + 2 csc (5 x + 1) + sin (5 x + 1)) + C

解答ステップ

\int \frac{cos ^{5} ( 5 x + 1 )}{sin ^{4} ( 5 x + 1 )} d x

u置換積分法を適用する

= \int \frac{cos ^{5} ( u )}{5 sin ^{4} ( u )} d u

定数を除く:

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

= \frac{1}{5} \cdot \int \frac{cos ^{5} ( u )}{sin ^{4} ( u )} d u

三角関数の公式を使用して書き換える

= \frac{1}{5} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) ^{2} cos ( u )}{sin ^{4} ( u )} d u

u置換積分法を適用する

= \frac{1}{5} \cdot \int \frac{( 1 - v ^{2} ) ^{2}}{v ^{4}} d v

拡張

\frac{( 1 - v ^{2} ) ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{2}{v ^{2}} + 1

= \frac{1}{5} \cdot \int \frac{1}{v ^{4}} - \frac{2}{v ^{2}} + 1 d v

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

= \frac{1}{5} (\int \frac{1}{v ^{4}} d v - \int \frac{2}{v ^{2}} d v + \int 1 d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

\int \frac{2}{v ^{2}} d v = - \frac{2}{v}

\int 1 d v = v

= \frac{1}{5} (- \frac{1}{3 v ^{3}} - (- \frac{2}{v}) + v)

代用を戻す

= \frac{1}{5} (- \frac{1}{3 sin ^{3} ( 5 x + 1 )} - (- \frac{2}{sin ( 5 x + 1 )}) + sin (5 x + 1))

簡素化

\frac{1}{5} (- \frac{1}{3 sin ^{3} ( 5 x + 1 )} - (- \frac{2}{sin ( 5 x + 1 )}) + sin (5 x + 1)) : \frac{1}{5} (- \frac{1}{3} csc^{3} (5 x + 1) + 2 csc (5 x + 1) + sin (5 x + 1))

= \frac{1}{5} (- \frac{1}{3} csc^{3} (5 x + 1) + 2 csc (5 x + 1) + sin (5 x + 1))

定数を解答に追加する

= \frac{1}{5} (- \frac{1}{3} csc^{3} (5 x + 1) + 2 csc (5 x + 1) + sin (5 x + 1)) + C

グラフ

人気の例

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