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

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

Popular >

integral of 1/(sqrt(4^2-x^2))

\int \frac{1}{4 ^{2} - x ^{2}} d x

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

Solution steps

\int \frac{1}{4 ^{2} - x ^{2}} d x

4^{2} - x^{2} = 4^{2} (1 - \frac{x ^{2}}{4 ^{2}})

= \int \frac{1}{4 ^{2} ( 1 - \frac{x ^{2}}{4 ^{2}} )} d x

Simplify

\frac{1}{4 ^{2} ( 1 - \frac{x ^{2}}{4 ^{2}} )} : \frac{1}{16 - x ^{2}}

= \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{t}{( t - 6 ) ^{2}}

x \to - 8 - lim (\frac{∣ x + 8 ∣}{x + 8})

n = 0 \sum \infty \frac{x ^{n}}{n + 1}

f (x) = (3 x + 1) (5 x - 4)

\frac{d}{d x} ((45 - 0.12 x) x)

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