What is the integral from 0 to pi/2 of tsin(t) ?

The integral from 0 to pi/2 of tsin(t) is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of tsin(t)

\int_{0}^{\frac{π}{2}} t sin (t) d t

1

Solution steps

\int_{0}^{\frac{π}{2}} t sin (t) d t

Apply Integration By Parts

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

= [- t cos (t) - \int - cos (t) d t]_{0}^{\frac{π}{2}}

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

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

Simplify

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

Compute the boundaries:

1

= 1

\int_{- 2}^{2} (1 - \frac{x ^{2}}{4})^{2} d x

\int_{0}^{2 x^{2}} 3 x y d y

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

\int_{0}^{3} (3 - x)^{\frac{- 1}{2}} d x

\int_{- 6}^{1} x^{3} d x

What is the integral from 0 to pi/2 of tsin(t) ?
The integral from 0 to pi/2 of tsin(t) is 1