What is the integral from 0 to pi/2 of 1/(x^2+2) ?

The integral from 0 to pi/2 of 1/(x^2+2) is (arctan(pi/(2sqrt(2))))/(sqrt(2))

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integral from 0 to pi/2 of 1/(x^2+2)

\int_{0}^{\frac{π}{2}} \frac{1}{x ^{2} + 2} d x

\frac{arctan ( \frac{π}{2 2} )}{2}

Decimal

0.59241 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{1}{x ^{2} + 2} d x

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} [arctan (u)]_{0}^{\frac{π}{2 2}}

Compute the boundaries:

arctan (\frac{π}{2 2})

= \frac{1}{2} arctan (\frac{π}{2 2})

Simplify

= \frac{arctan ( \frac{π}{2 2} )}{2}

x \to \infty lim (\frac{x ^{5}}{x ^{10} + 9})

f (x) = x^{2} sin (x) + 2 x cos (x)

\frac{d y}{d x} = x y + 2 y

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{4.5 e ^{5 t}}{tt + 1}

\frac{d y}{d x} = 6 x^{5} cos^{2} (y)

What is the integral from 0 to pi/2 of 1/(x^2+2) ?
The integral from 0 to pi/2 of 1/(x^2+2) is (arctan(pi/(2sqrt(2))))/(sqrt(2))