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

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

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integral of 1/(sqrt(1-4t^2))

\int \frac{1}{1 - 4 t ^{2}} d t

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

Solution steps

\int \frac{1}{1 - 4 t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsin (2 t)

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} = 0.2 ln (\frac{4000}{y}) y

\int_{0}^{1} 2

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

x \to 0 lim (x^{\frac{- 2}{3}})

f (x) = x^{3} - 3 x + 2, (2, 4)

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