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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

提出常数:

\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

使用换元积分法

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

提出常数:

\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

使用分布积分法

= 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))

u = 3 x

代回

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

化简

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))

解答补常数

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

作图

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