Integration by Parts

[Runner88]

Hello all,
First of many posts and just want thank yall for being here, I have a question regarding some integration by parts that may also just be an error in Khan Academy but I was spending a good 45 mins on this problem and thought it was right but when i entered it into the module it said I was wrong. Below is a screen shot of part in the problem where I guess my error was or theirs was but my question is, where does the 4/25 come from shouldnt it be a -2/5? I understand where the 2/25 comes from as marked in green but did they just make an error or am I just over tired.

Thank you

 

[Johulus]

\(\displaystyle \int \mathrm{uv'} \mathrm{d}x=uv- \int \mathrm{u'v} \mathrm{d}x \)

So, for \(\displaystyle u=\sin{(2x)} \) and \(\displaystyle v'=\cos{(5x)} \) we have \(\displaystyle u'=2\cos{(2x)} \) and \(\displaystyle v=\dfrac{1}{5} \sin{(5x)} \). And now you can plug it in the first equation.

\(\displaystyle \int \mathrm{\sin{(2x)} \cos{(5x)}} \mathrm{d}x=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \int \mathrm{2\cos{(2x)} \cdot \dfrac{1}{5}\sin{(5x)}} \mathrm{d}x=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \dfrac{2}{5}\int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x \)

Now let's just look at \(\displaystyle \int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x \) separately.

Let's say that now \(\displaystyle u=\cos{(2x)} \) and \(\displaystyle v'=\sin{(5x)} \). Then \(\displaystyle u'=-2\sin{(2x)} \) and \(\displaystyle v=-\dfrac{1}{5}\cos{(5x)} \).

Then we have \(\displaystyle \int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x= -\dfrac{1}{5}\cos{(2x)}\cos{(5x)}- \int \mathrm{-2\sin{(2x)} \cdot (-\dfrac{1}{5}\cos{(5x)})} \mathrm{d}x =-\dfrac{1}{5}\cos{(2x)}\cos{(5x)}- \dfrac{2}{5}\int \mathrm{\sin{(2x)}\cos{(5x)}} \mathrm{d}x\)

Now we have:
\(\displaystyle \int \mathrm{\sin{(2x)} \cos{(5x)}} \mathrm{d}x=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \int \mathrm{2\cos{(2x)} \cdot \dfrac{1}{5}\sin{(5x)}} \mathrm{d}x= \\ \dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \dfrac{2}{5}\int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x =\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \dfrac{2}{5}(-\dfrac{1}{5}\cos{(2x)}\cos{(5x)}- \\ \dfrac{2}{5}\int \mathrm{\sin{(2x)}\cos{(5x)}} \mathrm{d}x)=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}+ \dfrac{2}{25} \cos{(2x)}\cos{(5x)}+ \dfrac{4}{25}\int \mathrm{\sin{(2x)}\cos{(5x)}} \mathrm{d}x \)

Everything seems to be just fine.

 

[Runner88]

The 2/5 is distributed.....

\(\displaystyle \int \mathrm{uv'} \mathrm{d}x=uv- \int \mathrm{u'v} \mathrm{d}x \)

So, for \(\displaystyle u=\sin{(2x)} \) and \(\displaystyle v'=\cos{(5x)} \) we have \(\displaystyle u'=2\cos{(2x)} \) and \(\displaystyle v=\dfrac{1}{5} \sin{(5x)} \). And now you can plug it in the first equation.

\(\displaystyle \int \mathrm{\sin{(2x)} \cos{(5x)}} \mathrm{d}x=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \int \mathrm{2\cos{(2x)} \cdot \dfrac{1}{5}\sin{(5x)}} \mathrm{d}x=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \dfrac{2}{5}\int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x \)

Now let's just look at \(\displaystyle \int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x \) separately.

Let's say that now \(\displaystyle u=\cos{(2x)} \) and \(\displaystyle v'=\sin{(5x)} \). Then \(\displaystyle u'=-2\sin{(2x)} \) and \(\displaystyle v=-\dfrac{1}{5}\cos{(5x)} \).

Then we have \(\displaystyle \int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x= -\dfrac{1}{5}\cos{(2x)}\cos{(5x)}- \int \mathrm{-2\sin{(2x)} \cdot (-\dfrac{1}{5}\cos{(5x)})} \mathrm{d}x =-\dfrac{1}{5}\cos{(2x)}\cos{(5x)}- \dfrac{2}{5}\int \mathrm{\sin{(2x)}\cos{(5x)}} \mathrm{d}x\)

Now we have:
\(\displaystyle \int \mathrm{\sin{(2x)} \cos{(5x)}} \mathrm{d}x=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \int \mathrm{2\cos{(2x)} \cdot \dfrac{1}{5}\sin{(5x)}} \mathrm{d}x= \\ \dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \dfrac{2}{5}\int \mathrm{\cos{(2x)}\sin{(5x)}} \mathrm{d}x =\dfrac{1}{5}\sin{(2x)}\sin{(5x)}- \dfrac{2}{5}(-\dfrac{1}{5}\cos{(2x)}\cos{(5x)}- \\ \dfrac{2}{5}\int \mathrm{\sin{(2x)}\cos{(5x)}} \mathrm{d}x)=\dfrac{1}{5}\sin{(2x)}\sin{(5x)}+ \dfrac{2}{25} \cos{(2x)}\cos{(5x)}+ \dfrac{4}{25}\int \mathrm{\sin{(2x)}\cos{(5x)}} \mathrm{d}x \)

Everything seems to be just fine.

I must have been more tired than I thought, I wasn't seeing the distribution of the -2/5 to the -1/5 and then other -2/5

thank you