What is the integral of 2t(9-4t^2)^{1/2} ?

The integral of 2t(9-4t^2)^{1/2} is -1/6 (9-4t^2)^{3/2}+C

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integral of 2t(9-4t^2)^{1/2}

integral

2 t (9 - 4 t^{2})^{\frac{1}{2}}

- \frac{1}{6} (9 - 4 t^{2})^{\frac{3}{2}} + C

Solution steps

\int 2 t (9 - 4 t^{2})^{\frac{1}{2}} d t

Take the constant out:

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

= 2 \cdot \int t (9 - 4 t^{2})^{\frac{1}{2}} d t

Apply u-substitution

: \int - \frac{u ^{\frac{1}{2}}}{8} d u

= 2 \cdot \int - \frac{u ^{\frac{1}{2}}}{8} d u

Take the constant out:

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

= 2 (- \frac{1}{8} \cdot \int u^{\frac{1}{2}} d u)

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 2 (- \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 9 - 4 t^{2}

= 2 (- \frac{1}{8} \cdot \frac{2}{3} (9 - 4 t^{2})^{\frac{3}{2}})

Simplify

2 (- \frac{1}{8} \cdot \frac{2}{3} (9 - 4 t^{2})^{\frac{3}{2}}) : - \frac{1}{6} (9 - 4 t^{2})^{\frac{3}{2}}

= - \frac{1}{6} (9 - 4 t^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (9 - 4 t^{2})^{\frac{3}{2}} + C

- 4 x

x^{\frac{1}{6}}

\frac{1}{x ^{3} + 1}

\frac{2 y ^{2}}{( x + y ) ^{2}}

\frac{1}{e ^{3}} x + sin (x)

What is the integral of 2t(9-4t^2)^{1/2} ?
The integral of 2t(9-4t^2)^{1/2} is -1/6 (9-4t^2)^{3/2}+C