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

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

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

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

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

Solution steps

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

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} t)

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

Simplify

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

Add a constant to the solution

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

f (x) = ln (1 + x)

\frac{d}{d x} (ln (\frac{2 x}{sin ( x )}))

y = \frac{3}{sin ( 2 x )}

x \to 1 lim (\frac{( x + 3 )}{2})

\int (a x + b)^{n} d x

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