What is the integral from 0 to pi/2 of cos(sqrt(2x)) ?

The integral from 0 to pi/2 of cos(sqrt(2x)) is sqrt(pi)sin(sqrt(pi))+cos(sqrt(pi))-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 cos(sqrt(2x))

\int_{0}^{\frac{π}{2}} cos (2 x) d x

π sin (π) + cos (π) - 1

Decimal

0.53624 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos (2 x) d x

Apply u-substitution

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

= \int_{0}^{π} cos (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} cos (u) d u

Apply u-substitution

: \int_{0}^{π} cos (v) \cdot 2 v d v

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

Take the constant out:

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

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

Apply Integration By Parts

: [v sin (v) - \int sin (v) d v]_{0}^{π}

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

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

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

Simplify

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

= [v sin (v) + cos (v)]_{0}^{π}

Compute the boundaries:

π sin (π) + cos (π) - 1

= π sin (π) + cos (π) - 1

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

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

\int_{\frac{y}{2}}^{1} (y e^{x^{3}}) d x

\int_{1}^{4} \frac{18 ( ln ( x ) ) ^{5}}{x} d x

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

What is the integral from 0 to pi/2 of cos(sqrt(2x)) ?
The integral from 0 to pi/2 of cos(sqrt(2x)) is sqrt(pi)sin(sqrt(pi))+cos(sqrt(pi))-1