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

The integral of 1/(sqrt(16-x^2)) is arcsin(1/4 x)+C

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

\int \frac{1}{16 - x ^{2}} d x

arcsin (\frac{1}{4} x) + C

Solution steps

\int \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

= arcsin (\frac{1}{4} x) + C

{\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}}}

x \to 0 lim (\frac{x}{sin ( 4 ) ( x )})

\int e^{\frac{1}{2} t} d t

7 t^{7} + 9

x^{- m}

What is the integral of 1/(sqrt(16-x^2)) ?
The integral of 1/(sqrt(16-x^2)) is arcsin(1/4 x)+C