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人気のある微分積分 >

integral of 6sin^2(3x)cos(3x)ln(sin(3x))

解

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

解

2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) + C

解答ステップ

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

定数を除く:

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

= 6 \cdot \int sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

u置換積分法を適用する

= 6 \cdot \int sin^{2} (u) cos (u) ln (sin (u)) \frac{1}{3} d u

定数を除く:

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

= 6 \cdot \frac{1}{3} \cdot \int sin^{2} (u) cos (u) ln (sin (u)) d u

部分積分を適用する

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (u) ln (sin (u)) - \int \frac{1}{3} sin^{3} (u) cot (u) d u)

\int \frac{1}{3} sin^{3} (u) cot (u) d u = \frac{1}{9} sin^{3} (u)

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (u) ln (sin (u)) - \frac{1}{9} sin^{3} (u))

代用を戻す

u = 3 x

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

簡素化

6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) : 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

= 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

定数を解答に追加する

= 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) + C

グラフ

人気の例

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