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

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

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

arcsin (4 x) + C

Solution steps

\int \frac{4}{1 - 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}{1 - 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 (4 x)

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

Simplify

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

= arcsin (4 x)

Add a constant to the solution

= arcsin (4 x) + C

\int \frac{5}{x ( x ^{2} + 5 ) ^{2}} d x

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

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

\int 2 x^{3} sin (x^{2}) d x

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