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

The integral from 0 to pi/2 of (sin(2t)) 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 (sin(2t))

\int_{0}^{\frac{π}{2}} (sin (2 t)) d t

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 t) d t

Apply u-substitution

: \int_{0}^{π} sin (u) \frac{1}{2} d u

= \int_{0}^{π} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{2} \cdot 2

Simplify

= 1

\int_{0}^{1} 4 x + 3 d x

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

\int_{0}^{\frac{π}{4}} (2 sec^{2} (x)) d x

\int_{0}^{2 π} \frac{8}{3} d θ

\int_{0}^{t^{10}} u d u

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