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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{e ^{\frac{π}{2}} + 1}{2}

Decimal

2.90523 \dots

Solution steps

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

Apply Integration By Parts

= [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}}}

Simplify

[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}}}

Compute the boundaries:

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

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

Simplify

= \frac{e ^{\frac{π}{2}} + 1}{2}

What is the integral from 1 to e^{pi/2 of}sin(ln(θ)) ?
The integral from 1 to e^{pi/2 of}sin(ln(θ)) is (e^{pi/2}+1)/2