What is the integral from 4 to 5 of 1/(sqrt(25-x^2)) ?

The integral from 4 to 5 of 1/(sqrt(25-x^2)) is pi/2-arcsin(4/5)

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integral from 4 to 5 of 1/(sqrt(25-x^2))

\int_{4}^{5} \frac{1}{25 - x ^{2}} d x

\frac{π}{2} - arcsin (\frac{4}{5})

Decimal

0.64350 \dots

Solution steps

\int_{4}^{5} \frac{1}{25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{a r c s i n (\frac{4}{5})}^{\frac{π}{2}} 1 d u

= \int_{a r c s i n (\frac{4}{5})}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{a r c s i n (\frac{4}{5})}^{\frac{π}{2}}

Simplify

= [u]_{a r c s i n (\frac{4}{5})}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2} - arcsin (\frac{4}{5})

= \frac{π}{2} - arcsin (\frac{4}{5})

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What is the integral from 4 to 5 of 1/(sqrt(25-x^2)) ?
The integral from 4 to 5 of 1/(sqrt(25-x^2)) is pi/2-arcsin(4/5)