integrate the following equation (what am I doing wrong here 2)

Question

Here is the equation:

$$\int 3x \sqrt{1-2x^2}dt$$

Here is my answer:

$$ \dfrac14 \int (1-2x^2)^{1/2} . 3x = -\dfrac14 \dfrac{(1-2x^2)^{3/2}}{3/2} = -\dfrac14 \cdot \dfrac23 (1-2x^2)^{3/2} + c$$

correct answer:

it should be -1/2 instead of -1/4 . 2/3 at the end. what am I doing wrong?

Answer 1

let us denote $s=1-2*x^2$,then $ds=-4*xdx$ ,because you have $3$ in your equation,outside of integrat will come $-3/4$,so it would be

$$\int (-3/4)*\sqrt{s}ds$$

can you continue from this?

Answer 2

let $$\sqrt{1-2x^2}=t$$ then $$-4xdx=dt$$ the integral becomes $$\int \frac{-3}{4}\sqrt{t}dt$$ $$= \frac{-3}{4}.\frac{2}{3}t^\frac{3}{2}$$

Answer 3

Another way:

Denote $t^2 = 1-(\sqrt{2}x)^2$. Then you get $dx=-\frac{tdt}{\sqrt{2}x}$. So you get cancellations and the integral becomes $$ -\frac{3}{\sqrt{2}} \int |t|tdt= -\frac{t^3 \ sign (t)}{\sqrt{2}}+C $$ or just $-\frac{t^3}{\sqrt{2}}+C$ assuming $t>0$.