ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 1 to e^{pi/2 of}sin(ln(θ))

解

\int_{1}^{e^{\frac{π}{2}}} sin (ln (θ)) d θ

解

\frac{e ^{\frac{π}{2}} + 1}{2}

+1

十進法表記

2.90523 \dots

解答ステップ

\int_{1}^{e^{\frac{π}{2}}} sin (ln (θ)) d θ

部分積分を適用する

= [θ sin (ln (θ)) - \int cos (ln (θ)) d θ]_{1}^{e^{\frac{π}{2}}}

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

= [θ sin (ln (θ)) - (\frac{1}{2} θ sin (ln (θ)) + \frac{1}{2} θ cos (ln (θ)))]_{1}^{e^{\frac{π}{2}}}

簡素化

[θ sin (ln (θ)) - (\frac{1}{2} θ sin (ln (θ)) + \frac{1}{2} θ cos (ln (θ)))]_{1}^{e^{\frac{π}{2}}} : [\frac{1}{2} (θ sin (ln (θ)) - θ cos (ln (θ)))]_{1}^{e^{\frac{π}{2}}}

= [\frac{1}{2} (θ sin (ln (θ)) - θ cos (ln (θ)))]_{1}^{e^{\frac{π}{2}}}

限度を計算する:

\frac{e ^{\frac{π}{2}}}{2} + \frac{1}{2}

= \frac{e ^{\frac{π}{2}}}{2} + \frac{1}{2}

簡素化

= \frac{e ^{\frac{π}{2}} + 1}{2}

グラフ

人気の例

integral from-1 to 1 of x^2+x

\int_{- 1}^{1} x^{2} + x d x

integral from 0 to 2 of x^2(2-x)

\int_{0}^{2} x^{2} (2 - x) d x

integral from 0 to 3 of [x]

\int_{0}^{3} [x] d x

integral from 1 to 3 of x^2-x+1

\int_{1}^{3} x^{2} - x + 1 d x

integral from 0 to 1 of 1/(x^{1/4)}

\int_{0}^{1} \frac{1}{x ^{\frac{1}{4}}} d x