What is the integral from 0 to 1/2 of 5tcos(pit) ?

The integral from 0 to 1/2 of 5tcos(pit) is 5(1/(2pi)-1/(pi^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1/2 of 5tcos(pit)

\int_{0}^{\frac{1}{2}} 5 t cos (π t) d t

5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

Decimal

0.28916 \dots

Solution steps

\int_{0}^{\frac{1}{2}} 5 t cos (π t) d t

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{1}{2}} t cos (π t) d t

Apply Integration By Parts

: [\frac{1}{π} (t sin (π t) - π \cdot \int \frac{1}{π} sin (π t) d t)]_{0}^{\frac{1}{2}}

= 5 [\frac{1}{π} (t sin (π t) - π \cdot \int \frac{1}{π} sin (π t) d t)]_{0}^{\frac{1}{2}}

\int \frac{1}{π} sin (π t) d t = - \frac{1}{π ^{2}} cos (π t)

= 5 [\frac{1}{π} (t sin (π t) - π (- \frac{1}{π ^{2}} cos (π t)))]_{0}^{\frac{1}{2}}

Simplify

5 [\frac{1}{π} (t sin (π t) - π (- \frac{1}{π ^{2}} cos (π t)))]_{0}^{\frac{1}{2}} : 5 [\frac{1}{π} (t sin (π t) + \frac{1}{π} cos (π t))]_{0}^{\frac{1}{2}}

= 5 [\frac{1}{π} (t sin (π t) + \frac{1}{π} cos (π t))]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{2 π} - \frac{1}{π ^{2}}

= 5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

\int_{0}^{1} 3 x + 3 d x

\int_{3}^{2} sin (t) d t

\int_{- 2 π}^{2 π} 6 sin (x) d x

\int_{2}^{4} x^{3} - 2 - ln (x + 1) d x

\int_{\frac{1}{2}}^{5} 2 x - 1 d x

What is the integral from 0 to 1/2 of 5tcos(pit) ?
The integral from 0 to 1/2 of 5tcos(pit) is 5(1/(2pi)-1/(pi^2))