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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

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}

\int_{1}^{3} \frac{1}{x ^{4}} d x

\int_{0}^{1} 5 x + 6 d x

\int_{1}^{4} 2 x^{2} d x

\int_{0}^{8} 8 x d x

\int_{0}^{3} 5^{3 x} d x

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