What is the integral from 0 to 5 of 1/(sqrt(25-x^2)) ?

The integral from 0 to 5 of 1/(sqrt(25-x^2)) is pi/2

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integral from 0 to 5 of 1/(sqrt(25-x^2))

\int_{0}^{5} \frac{1}{25 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{5} \frac{1}{25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 1 d u

= \int_{0}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{2}}

Simplify

= [u]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\frac{\partial}{\partial y} (\frac{x ^{2}}{y ^{3}})

\frac{\partial}{\partial θ} (r (θ) cos^{2} (θ))

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

\frac{d}{d x} (A x e^{- 3 x})

\int \frac{csc ( θ ) cot ( θ )}{5 - 4 csc ( θ )} d θ

What is the integral from 0 to 5 of 1/(sqrt(25-x^2)) ?
The integral from 0 to 5 of 1/(sqrt(25-x^2)) is pi/2