What is the integral of 8/(sqrt(4-x^2)) ?

The integral of 8/(sqrt(4-x^2)) is 8arcsin(1/2 x)+C

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integral of 8/(sqrt(4-x^2))

\int \frac{8}{4 - x ^{2}} d x

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

Solution steps

\int \frac{8}{4 - x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 8 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 8 \cdot 1 \cdot u

Substitute back

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

= 8 \cdot 1 \cdot arcsin (\frac{1}{2} x)

Simplify

= 8 arcsin (\frac{1}{2} x)

Add a constant to the solution

= 8 arcsin (\frac{1}{2} x) + C

\int \frac{x ^{2} - 1}{x ^{3} - 3 x + 1} d x

cos^{s i n (x)} (x)

f (x) = e^{- 0.5 x} (4 - x) - 2

n = 1 \sum \infty \frac{5}{4 n ^{2} - 1}

\frac{d}{d x} (- 3 sin (6 x))

What is the integral of 8/(sqrt(4-x^2)) ?
The integral of 8/(sqrt(4-x^2)) is 8arcsin(1/2 x)+C