Melody

What is the bar at the front for?

A quadrant can also just be a quarter of any circle :)

This is what Alan is describing

$${\mathtt{360}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right) = {\mathtt{600}}$$     degrees

Thanks Alan,       I should have thought of that    

Thanks Heureka :)

That is fabulous Brittany  

51 is an excellent number because it is 3*17

And yes zegroes, at some point in my life I will be 51 :))

$${\sqrt{{\frac{{\mathtt{10}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{60}}^\circ\right)}}}}} = {\mathtt{3.398\: \!088\: \!489\: \!695\: \!106}}$$

Works fo me :)

Now it is right     I had just made a careless error right at the beginning.

Thanks for spotting it Alan     [ I do have a question at the bottom though]

the correct answer is   

I have not found my errors  :(

---------------------------------------------------------------------------------

$$\\\int\frac{(2cos3x + 3sinx) }{sin^3x }dx\\\\
=\int\frac{2(cos2xcosx-sin2xsinx) + 3sinx }{sin^3x }dx\\\\
=\int\frac{2cos2xcosx - 2sin2xsinx) + 3sinx }{sin^3x }dx\\\\
=\int\frac{2(cos^2x-sin^2x)cosx - 2(2sinxcosx)sinx) + 3sinx }{sin^3x }dx\\\\
=\int\frac{2cosxcos^2x-2cosxsin^2x - 4sin^2xcosx + 3sinx }{sin^3x }dx\\\\
=\int\frac{2cos^3x-2cosxsin^2x - 4sin^2xcosx + 3sinx }{sin^3x }dx\\\\
=\int\frac{2cos^3x - 6sin^2xcosx + 3sinx }{sin^3x }dx\\\\
=\int\;2cot^3x-\frac{6cosx}{sinx} + 3csc^2x \;dx\\\\
=\int\;2cot^3x\;dx -\int\;\frac{6cosx}{sinx}\;dx +\int\; 3csc^2x \;dx\\\\
=\int\;2cot^3x\;dx -6ln(sinx) +\int\; 3csc^2x \;dx\\\\
=-6ln(sinx)\;+\;\int\;2cot^3x\;dx -3cotx\\\\
=-6ln(sinx)\;-3cotx\;+\;\int\;2cot^3x\;dx$$

$$\\$NOW I'll use the reduction formula$\\\\
\int\;2cot^3x\;dx\\\\
=2[\frac{-Cot^{3-1}x}{3-1}-\int\;cot^{-2+3}x\;dx]\\\\
=2[\frac{-Cot^{2}x}{2}-\int\;cotx\;dx]\\\\
=-Cot^{2}x-2\;ln(sinx)\\\\$$

$$\\$So - continuing from before$\\\\
=-6ln(sinx)\;-3cotx\;+\;\int\;2cot^3x\;dx \\\\
=-6ln(sinx)\;-3cotx\;+\; -Cot^{2}x-2\;ln(sinx) +c \\\\
=-8ln(sinx)\;-3Cotx-Cot^{2}x \;+c \\\\$$

-------------------------------------------------------------

Now this is correct but Wolfram Alpha is telling me that  $$cot^2x=csc^2x$$   for the restricted values involed here.

WHY IS THAT ?

hi Bhustancrowe

Mmm

I spent ages doing it the wrong way, here is the right way.  I can't take the credit.  :(

Oh I managed to delete the bit i did correctly - this is the second one - you should

be able to do the first little bit of it  yourself   

It sounds like you had a really amazing day Brittany!

How many more days are there?