What is the integral from 0 to 1 of 1/(sqrt(16-x^2)) ?

The integral from 0 to 1 of 1/(sqrt(16-x^2)) is arcsin(1/4)

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integral from 0 to 1 of 1/(sqrt(16-x^2))

\int_{0}^{1} \frac{1}{16 - x ^{2}} d x

arcsin (\frac{1}{4})

Decimal Notation

0.25268 \dots

Solution steps

\int_{0}^{1} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{1}{4})} 1 d u

= \int_{0}^{a r c s i n (\frac{1}{4})} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{a r c s i n (\frac{1}{4})}

Simplify

= [u]_{0}^{a r c s i n (\frac{1}{4})}

Compute the boundaries:

arcsin (\frac{1}{4})

= arcsin (\frac{1}{4})

\int \frac{4 x ln ( 1 + x ^{2} )}{1 + x ^{2}} d x

y d x + x^{2} + 1 d y = 0

\int x x - 18 d x

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

(- 2.2) (2.3)

What is the integral from 0 to 1 of 1/(sqrt(16-x^2)) ?
The integral from 0 to 1 of 1/(sqrt(16-x^2)) is arcsin(1/4)