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

Here is the equation:

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

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

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

• You missed a factor of 3. Try to derivative your result and see where's missing. – Shuchang Dec 7 '13 at 16:47
• It should be $dx$, not $dt$. – egreg Dec 7 '13 at 16:52
• It is not an equation. – JP McCarthy Dec 9 '13 at 14:49

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?

• so it would be 12x.... – Cash Vai Dec 7 '13 at 16:50
• no look please ,i have updated – dato datuashvili Dec 7 '13 at 16:51
• you are welcome, good lucks – dato datuashvili Dec 7 '13 at 16:56
• @CashVai any question ?if yes you are welcome – dato datuashvili Dec 7 '13 at 17:02
• thank you very much! I will, stay tuned :DD – Cash Vai Dec 7 '13 at 17:11

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

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 $t>0$.