# Thread: another integration - need help!

1. ## another integration - need help!

8 times the integral ( from -1/sqrt(2) to 1/sqrt(2) ) of ( 1-2x^2)^3/2 dx

2. A trig. substitution would work... even is not a cool integral...

3. $\displaystyle \int\limits_{ - \frac{1}{{\sqrt 2 }}}^{\frac{1}{{\sqrt 2 }}} {\sqrt[3]{{\left( {1 - 2x^2 } \right)^2 }}} = 2\int\limits_0^{\frac{1}{{\sqrt 2 }}} {\sqrt[3]{{\left( {1 - 2x^2 } \right)^2 }}}$

4. Hi plato,

And then...

Could you please show me the full steps ? Thank you very much.

Hi Krizalid,

Please show me the full steps with trig substitution. I still can't work to the end with it. I really forgot the trig substitution work .

The previous question. I worked it out simply by using the face that the integral is equal to 1/2 pi a^2 .

5. Hello, kittycat!

$\displaystyle 8\int^{\frac{1}{\sqrt{2}}}_{-\frac{1}{\sqrt{2}}}\left(1-2x^2\right)^{\frac{3}{2}}\,dx$

Let: $\displaystyle 2x^2 \:=\:\sin^2\!\theta\quad\Rightarrow\quad x \:=\:\frac{1}{\sqrt{2}}\sin\theta\quad\Rightarrow\ quad dx \:=\:\frac{1}{\sqrt{2}}\cos\theta\,d\theta$
. . and: $\displaystyle \left(1 - 2x^2\right)^{\frac{3}{2}} \:=\:\left(1-\sin^2\!\theta\right)^{\frac{3}{2}} \:=\:\left(\cos^2\!\theta\right)^{\frac{3}{2}} \:=\:\cos^3\!\theta$

Substitute: .$\displaystyle 8\int \cos^3\!\theta\left(\frac{1}{\sqrt{2}}\cos\theta\, d\theta\right) \:=\:4\sqrt{2}\int\cos^4\!\theta\,d\theta \;=\;4\sqrt{2}\int\left(\cos^2\!\theta\right)^2d\t heta$

. . $\displaystyle = \;4\sqrt{2}\int\left(\frac{1 + \cos2\theta}{2}\right)^2d\theta \;=\;\sqrt{2}\int\left(1 + 2\cos2\theta + \cos^2\!2\theta\right)\,d\theta$

. . $\displaystyle = \;\sqrt{2}\int\left(1 + 2\cos2\theta + \frac{1 + \cos4\theta}{2}\right)d\theta \;= \;\sqrt{2}\int\left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right)d\theta$

Can you finish it now?

6. As you can see kittycat, Soroban gave you a full trig. sub. that I was talkin' about

7. Hi soroban,

Thank you very much for your detailed work.

I followed it but can't obtained the answer 3pi/2 ( the answer given by my teacher) . Do you get the final answer 3pi/2?

8. Hello, kittycat!

I don't get that answer either . . .

We have: .$\displaystyle \sqrt{2}\int \left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right)\,d\theta \;= \;\sqrt{2}\left(\frac{3}{2}\theta + \sin2\theta + \frac{1}{8}\sin4\theta\right)$

. . . . . $\displaystyle = \;\sqrt{2}\left[\frac{3}{2}\theta + 2\sin\theta\cos\theta + \frac{1}{4}\sin\theta\cos\theta(1 - 2\sin^2\theta)\right]$

Back-substitute: .$\displaystyle \sqrt{2}\left[\frac{3}{2}\arcsin(\sqrt{2}x) + 2\sqrt{2}x\sqrt{1-2x^2} + \frac{1}{4}\sqrt{2}x\sqrt{1-2x^2}(1-4x^2)\right]^{\frac{1}{\sqrt{2}}} _{\text{-}\frac{1}{\sqrt{2}}}$

Evaluate: .$\displaystyle \sqrt{2}\left[\left(\frac{3}{2}\arcsin(1) + 2\!\cdot\!0 + \frac{1}{4}\!\cdot\!0\right) - \left(\frac{3}{2}\arcsin(-1) + 2\!\cdot\!0 + \frac{1}{4}\!\cdot\!0\right)\right]$

. . . . . $\displaystyle = \;\sqrt{2}\left[\frac{3}{2}\!\cdot\!\frac{\pi}{2} - \frac{3}{2}\left(\text{-}\frac{\pi}{2}\right)\right] \;=\;\sqrt{2}\left[\frac{3\pi}{4} + \frac{3\pi}{4}\right] \;=\;\sqrt{2}\!\cdot\!\frac{3\pi}{2}$

So I get: .$\displaystyle \frac{3\sqrt{2}\pi}{2}\; \text{ or }\;\frac{3\pi}{\sqrt{2}}$

9. Hi soroban,

May I ask you one more question for this question? Why let 2x^2 =sin^2 theta ? Thank you very much.

I know in trig substitution, when you see sqrt ( a^2 -x^2) , you will let x = asintheta.