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受欢迎的微积分 >

integral of sin(10x)ln(sin(10x))

解答

\int sin (10 x) ln (sin (10 x)) d x

解答

\frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x)) + C

求解步骤

\int sin (10 x) ln (sin (10 x)) d x

使用换元积分法

= \int sin (u) ln (sin (u)) \frac{1}{10} d u

提出常数:

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

= \frac{1}{10} \cdot \int sin (u) ln (sin (u)) d u

使用分布积分法

= \frac{1}{10} (- ln (sin (u)) cos (u) - \int - cot (u) cos (u) d u)

\int - cot (u) cos (u) d u = - ln tan (\frac{u}{2}) - cos (u)

= \frac{1}{10} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

u = 10 x

代回

= \frac{1}{10} (- ln (sin (10 x)) cos (10 x) - (- ln tan (\frac{10 x}{2}) - cos (10 x)))

化简

\frac{1}{10} (- ln (sin (10 x)) cos (10 x) - (- ln tan (\frac{10 x}{2}) - cos (10 x))) : \frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x))

= \frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x))

解答补常数

= \frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x)) + C

作图

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