# Thread: [SOLVED] when integrating by method of substitution for a definite integral,answer di

1. ## [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...

2. Originally Posted by Aerospank
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... *
$\displaystyle \int_0^{\frac{2\pi}{3}}\sin^2(x)\cos(x)$. Let $\displaystyle z=\sin(x)\implies dz=\cos(x) dx$. So then our integral is equivalent to $\displaystyle \int_{\sin(0)}^{\sin\left(\frac{2\pi}{3}\right)}z^ 2\text{ }dz=\frac{\sqrt{3}}{8}$.

3. Originally Posted by Drexel28
$\displaystyle \int_0^{\frac{2\pi}{3}}\sin^2(x)\cos(x)$. Let $\displaystyle z=\sin(x)\implies dz=\cos(x) dx$. So then our integral is equivalent to $\displaystyle \int_{\sin(0)}^{\sin\left(\frac{2\pi}{3}\right)}z^ 2\text{ }dz=\frac{\sqrt{3}}{8}$.
That is incorrect...this is the original question:
Mathass.jpg picture by Aerospank - Photobucket

4. Originally Posted by Aerospank
That is incorrect...this is the original question:
Mathass.jpg picture by Aerospank - Photobucket

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.

5. Originally Posted by Defunkt
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.
Thank you so much. I will be careful in posting the correct question next time.

To Drexel: Sorry..and thanks for your kindness to help.

6. Originally Posted by Defunkt

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 $\displaystyle u(1-u^2)$.
ah, the result is $\displaystyle u\sqrt{1-u^2}$ because if you think of it, du = cosx dx. therefore dx = du/cosx. When substituting only this 'dx' expression in the integral at first, one of the cosx cancels out.

7. 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.