What is the integral of 4/(sqrt(64-16x^2)) ?

The integral of 4/(sqrt(64-16x^2)) is arcsin(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4/(sqrt(64-16x^2))

\int \frac{4}{64 - 16 x ^{2}} d x

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

Solution steps

\int \frac{4}{64 - 16 x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{64 - 16 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= 4 \cdot \int \frac{1}{4} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{4}{8} x)

= 4 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{8} x)

Simplify

4 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{8} x) : arcsin (\frac{x}{2})

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

Add a constant to the solution

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

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

\int \frac{x ^{3} + x ^{2} + 2 x + 3}{x ^{4} + 5 x ^{2} + 6} d x

\int \frac{2 x + 1}{x ^{2} - 4 x + 5} d x

\int \frac{e ^{x}}{e ^{2 x} - 2 e ^{x} + 5} d x

\int (x^{5} + 7) d x

What is the integral of 4/(sqrt(64-16x^2)) ?
The integral of 4/(sqrt(64-16x^2)) is arcsin(x/2)+C