Melody

No, it wasn't a trick Heureka

$$\int x\sqrt{0.25-x^2}\;dx\\\\
=\int x(0.25-x^2)^{1/2}\;dx\\\\
$Now I noted that $\frac{d}{dx}(0.25-x^2)=-2x \;\;$So the answer is going to be a multiple of $(0.25-x^2)^{1.5}\\\\\\
\frac{d}{dx}(0.25-x^2)^{1.5}\\\\
=\frac{3}{2}(0.25-x^2)^{0.5}*-2x\\\\
=\frac{3}{2}*-2x(0.25-x^2)^{0.5}\\\\\\
$so going backwards$\\\\
=\int \frac{3}{2}*-2x(0.25-x^2)^{1.5}dx\\\\
=x\sqrt{0.25-x^2}+c\\\\$$

$$\\so\\\\
\int \frac{3}{2}*-2x(0.25-x^2)^{1.5}dx=x(0.25-x^2)^{0.5}+c\\\\
\int \frac{2}{3}*\frac{-1}{2}*\frac{3}{2}*-2x(0.25-x^2)^{1.5}dx=\frac{2}{3}*\frac{-1}{2}*(0.25-x^2)^{0.5}+c\\\\
\int x(0.25-x^2)^{1.5}dx=\frac{2}{3}*\frac{-1}{2}*(0.25-x^2)^{0.5}+c\\\\
\int x(0.25-x^2)^{1.5}dx=\frac{-1}{3}*(0.25-x^2)^{0.5}+c\\\\$$

I didn't need to do all that.  i just took the reciprocal of the -2 and the 3/2 in the first place

$$\frac{-1}{2}\times \frac{2}{3}=-\frac{1}{3}$$

Sorry I took so long I was having a hard time explainining.  

I kind of 'stumble' my way through these.  It was good that you made me think about it properly   

(I hope I didn't stuff up )

If i had to do it your way I'd never have gotten it out!  

It is usually me that has the reputation for doing it the LONG way.  Looks like it was your turn :))

$$\\\frac{-(0.25-x^2)^{3/2}}{3}+c\\\\
=\frac{-(\frac{1-4x^2}{4})^{3/2}}{3}+c\\\\
=\frac{-\left(\frac{(1-4x^2)^{3/2}}{4^{3/2}}\right)}{3}+c\\\\
=-\left(\frac{(1-4x^2)^{3/2}}{8}\right)\times \frac{1}{3}+c\\\\
=-\frac{(1-4x^2)^{3/2}}{24}+c\\\\
=-\frac{1}{24}\;(1-4x^2)^{3/2}+c\\\\
=\;$Heureka's answer :)$$$

Heureka, do you think my answer is wrong?  They are probably both  the same 

 |-6| + 3|4| + |-5| - 0 · 6.

= +6+ 3*4  +   5     - 0

= +6+ 12  +   5 

=   23           

I have a question.

If you get rid of an irrational number in the denominator.  Then you are rationalizing the denominator.

If you get rid of an imaginary number in the denominator does that mean you are realizing the denominator?

It sounds funny but it is a real question :))

Thanks Alan and Heureka :)

Heureka has worked out the answer if j is an imaginary number   $$j=\sqrt{-1}$$               

Sometimes j is used instead of i to mean  $$\sqrt{-1}$$

cot(-68)=-cot(68)

this is because cot is 1/tan and tan is negative in the 4th quadrant.

Thanks Chris,

These are always great logic questions and this one is basic enough for our younger members to try to understand    

$${\frac{{\mathtt{0.018\: \!869}}}{{\mathtt{0.037\: \!6}}}} = {\mathtt{0.501\: \!835\: \!106\: \!382\: \!978\: \!7}}$$        

Here are some ideas

* Whenever you use a formula write it doen from memory.  Do this even if it is already committed to memory.   It will help it to stay there.

* If you do not know it then look it up. BUT never write it into your current question while looking at it elsewhere.  You must commit it to memory for at least a couple of seconds or you are not permitted to use it at all!    Write it on a scrap of paper a few times, it this is impossible at the beginning.

Make the time gap bigger and bigger and eventually write it and THEN if there is any doubt, check it is correct before using it.

* I am really bad at remembering formula, I often get around this by remembering (and understanding ) the derivation for it.  I often find it much easier to remember a long logical derivation then I do remembering the actual formula that it leads to.

These are just ideas.  Listen to everyone then do what works for you. Last time someone posted a question like this one of the mathematicians said that they wrote the formula down hundreds of times till it was committed to memory :)

I nearly forgot.  You can use poems and songs etc to help you as well.  

One you may already use is sohcahtoa    :))