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人気のある微分積分 >

integral of 1/((2+cos(x))sin(x))

解

\int \frac{1}{( 2 + cos ( x ) ) sin ( x )} d x

解

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

解答ステップ

\int \frac{1}{( 2 + cos ( x ) ) sin ( x )} d x

u置換積分法を適用する

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

拡張

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

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

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

\int \frac{1}{u ( u ^{2} + 3 )} d u = \frac{1}{3} ln ∣ u ∣ - \frac{1}{6} ln u^{2} + 3

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

= \frac{1}{3} ln ∣ u ∣ - \frac{1}{6} ln u^{2} + 3 + \frac{1}{2} ln u^{2} + 3

代用を戻す

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

= \frac{1}{3} ln tan (\frac{x}{2}) - \frac{1}{6} ln tan^{2} (\frac{x}{2}) + 3 + \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 3

類似した元を足す:

- \frac{1}{6} ln tan^{2} (\frac{x}{2}) + 3 + \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 3 = \frac{1}{3} ln tan^{2} (\frac{x}{2}) + 3

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

定数を解答に追加する

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

グラフ

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