What is the integral from 0 to pi/4 of cos(sqrt(x)) ?

The integral from 0 to pi/4 of cos(sqrt(x)) is sqrt(pi)sin((sqrt(pi))/2)+2cos((sqrt(pi))/2)-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 pi/4 of cos(sqrt(x))

\int_{0}^{\frac{π}{4}} cos (x) d x

π sin (\frac{π}{2}) + 2 cos (\frac{π}{2}) - 2

Decimal

0.63778 \dots

Solution steps

\int_{0}^{\frac{π}{4}} cos (x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

\frac{π}{2} sin (\frac{π}{2}) + cos (\frac{π}{2}) - 1

= 2 (\frac{π}{2} sin (\frac{π}{2}) + cos (\frac{π}{2}) - 1)

Simplify

= π sin (\frac{π}{2}) + 2 cos (\frac{π}{2}) - 2

\int_{- 2}^{2} x (x^{2} - 7) d x

\int_{0}^{π} e^{x} cos^{2} (x) d x

\int_{1}^{e} \frac{3}{x} d x

\int_{0}^{\infty} t^{3} e^{- t} d t

\int_{0}^{1} \frac{e ^{z} + 6}{e ^{z} + 6 z} d z

What is the integral from 0 to pi/4 of cos(sqrt(x)) ?
The integral from 0 to pi/4 of cos(sqrt(x)) is sqrt(pi)sin((sqrt(pi))/2)+2cos((sqrt(pi))/2)-2