I am integrating sin(squared)x cosx. With upper limit 2pi/3 and lower limit 0.

i get Answer = 3/8.

But when I substitute u = sinx, i get answer = 7/24.

Why is this?

I don't think I made any mistakes...

Printable View

- Jan 1st 2010, 11:39 AMAerospank[SOLVED] when integrating by method of substitution for a definite integral,answer di
I am integrating sin(squared)x cosx. With upper limit 2pi/3 and lower limit 0.

i get Answer = 3/8.

But when I substitute u = sinx, i get answer = 7/24.

Why is this?

I don't think I made any mistakes... - Jan 1st 2010, 11:43 AMDrexel28
- Jan 1st 2010, 12:42 PMAerospank
That is incorrect...this is the original question:

Mathass.jpg picture by Aerospank - Photobucket

Thanks in advance! - Jan 1st 2010, 01:15 PMDefunkt
No, actually, Drexel is correct:

$\displaystyle \int_{0}^{sin(\frac{2 \pi }{3})} z^2 ~ dz = \int_{0}^{\frac{\sqrt{3}}{2}} z^2 ~ dz = (\frac{\sqrt{3}}{2})^3 \cdot \frac{1}{3}$ $\displaystyle = \frac{3 \cdot \sqrt{3}}{8} \cdot \frac{1}{3} = \frac{\sqrt{3}}{8}$

Perhaps next time you should bother to see that you're actually posting the correct integral (you posted $\displaystyle \int sin^2(x)cos(x)$ while it is, in your page, $\displaystyle \int cos^2(x)sin(x)$)...

Also, it is not necessarily correct that $\displaystyle cosx = \sqrt{1-sin^2x}$ since, depending on the angle, it could also be that $\displaystyle cosx = -\sqrt{1-sin^2x}$ - the first is correct for $\displaystyle 2(k-1)\pi \leq x \leq \frac{k}{2}\pi, ~\frac{3k}{4}\pi \leq x \leq 2k\pi$ while the second is correct for $\displaystyle k\pi \leq x \leq \frac{3k}{4}\pi$.

This is apparently the mistake they wanted you to find out in that excersize (I really can't figure out if this is it since there are so many mistakes regardless).

Also, after claiming that $\displaystyle cos(x) = \sqrt{1-sin^2x}= \sqrt{1-u^2}$ they substitute it back into $\displaystyle cos^2x\cdot sinx$ as $\displaystyle u\sqrt{1-u^2}$ while if the first claim were true, the substitution would result in $\displaystyle u(1-u^2)$.

So there are two mistakes here. I can't tell if they are intentional or not (this text seems very low quality), but it seems as if they wanted you to note that the first fact I pointed out is what causes the different result.

This one is another great example of why you should post the whole question and not just parts of it. - Jan 1st 2010, 01:32 PMAerospank
- Jan 1st 2010, 08:34 PMAerospank
- Jan 2nd 2010, 04:41 AMDefunkt
That is correct, my bad. However, the mistake is where they used $\displaystyle cosx=\sqrt{1-sin^2x}$.

As you can see here, using $\displaystyle cosx = -\sqrt{1-sin^2x}$ in the second quadrant gives us the correct result.