integral of sin(ln(t))

解

\int sin (ln (t)) d t

\frac{1}{2} t sin (ln (t)) - \frac{1}{2} t cos (ln (t)) + C

解答ステップ

\int sin (ln (t)) d t

部分積分を適用する

= t sin (ln (t)) - \int cos (ln (t)) d t

\int cos (ln (t)) d t = \frac{1}{2} t sin (ln (t)) + \frac{1}{2} t cos (ln (t))

= t sin (ln (t)) - (\frac{1}{2} t sin (ln (t)) + \frac{1}{2} t cos (ln (t)))

簡素化

t sin (ln (t)) - (\frac{1}{2} t sin (ln (t)) + \frac{1}{2} t cos (ln (t))) : \frac{1}{2} t sin (ln (t)) - \frac{1}{2} t cos (ln (t))

= \frac{1}{2} t sin (ln (t)) - \frac{1}{2} t cos (ln (t))

定数を解答に追加する

= \frac{1}{2} t sin (ln (t)) - \frac{1}{2} t cos (ln (t)) + C