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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

: \int 1 d u

= 2 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot u

Substitute back

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

= 2 \cdot 1 \cdot arcsin (\frac{1}{4} x)

Simplify

= 2 arcsin (\frac{1}{4} x)

Add a constant to the solution

= 2 arcsin (\frac{1}{4} x) + C

\int - \frac{2}{81 - 49 x ^{2}} d x

\int \frac{3 x ^{2}}{( 49 + x ^{2} ) ^{2}} d x

\int a^{k x} d x

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

\int 0.024 \cdot (0.03 q^{2} - 2 q) + 130 d q

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