What is the integral of (1-4t^2)^{3/2} ?

The integral of (1-4t^2)^{3/2} is t(1-4t^2)^{3/2}+3/16 arcsin(2t)-3/64 sin(4arcsin(2t))+C

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integral of (1-4t^2)^{3/2}

\int (1 - 4 t^{2})^{\frac{3}{2}} d t

t (1 - 4 t^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 t) - \frac{3}{64} sin (4 arcsin (2 t)) + C

Solution steps

\int (1 - 4 t^{2})^{\frac{3}{2}} d t

Apply Integral Substitution

= \int \frac{( 1 - u ^{2} ) ^{\frac{3}{2}}}{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 - u^{2})^{\frac{3}{2}} d u

Apply Integration By Parts

: u (1 - u^{2})^{\frac{3}{2}} - \int - 3 u^{2} (1 - u^{2})^{\frac{1}{2}} d u

= \frac{1}{2} (u (1 - u^{2})^{\frac{3}{2}} - \int - 3 u^{2} (1 - u^{2})^{\frac{1}{2}} d u)

\int - 3 u^{2} (1 - u^{2})^{\frac{1}{2}} d u = - \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

= \frac{1}{2} (u (1 - u^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))))

Substitute back

u = 2 t

= \frac{1}{2} (2 t (1 - (2 t)^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (2 t) - \frac{1}{4} sin (4 arcsin (2 t)))))

Simplify

\frac{1}{2} (2 t (1 - (2 t)^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (2 t) - \frac{1}{4} sin (4 arcsin (2 t))))) : t (1 - 4 t^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 t) - \frac{3}{64} sin (4 arcsin (2 t))

= t (1 - 4 t^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 t) - \frac{3}{64} sin (4 arcsin (2 t))

Add a constant to the solution

= t (1 - 4 t^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 t) - \frac{3}{64} sin (4 arcsin (2 t)) + C

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What is the integral of (1-4t^2)^{3/2} ?
The integral of (1-4t^2)^{3/2} is t(1-4t^2)^{3/2}+3/16 arcsin(2t)-3/64 sin(4arcsin(2t))+C