What is the integral of (5t)/(sqrt(1-25t^4)) ?

The integral of (5t)/(sqrt(1-25t^4)) is 1/2 arcsin(5t^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5t)/(sqrt(1-25t^4))

\int \frac{5 t}{1 - 25 t ^{4}} d t

\frac{1}{2} arcsin (5 t^{2}) + C

Solution steps

\int \frac{5 t}{1 - 25 t ^{4}} d t

Take the constant out:

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

= 5 \cdot \int \frac{t}{1 - 25 t ^{4}} d t

Apply u-substitution

: \int \frac{1}{2 1 - 25 u ^{2}} d u

= 5 \cdot \int \frac{1}{2 1 - 25 u ^{2}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - 25 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{5} d v

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{5} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot v

Substitute back

= 5 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot arcsin (5 t^{2})

Simplify

5 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot arcsin (5 t^{2}) : \frac{1}{2} arcsin (5 t^{2})

= \frac{1}{2} arcsin (5 t^{2})

Add a constant to the solution

= \frac{1}{2} arcsin (5 t^{2}) + C

cot (csc (x))

g (x) = \frac{1}{x + 4}

\int_{1}^{5} \frac{1}{x - 1} d x

\frac{\partial}{\partial x} (e^{x \cdot e^{y}})

ln (1 - 3 x)

What is the integral of (5t)/(sqrt(1-25t^4)) ?
The integral of (5t)/(sqrt(1-25t^4)) is 1/2 arcsin(5t^2)+C