Hey looking for help ASAP have a test really soon, cant understand this problem. Will post below, calculus related (integrals). Medal ofcourse for answer, thanks a lot and if it can be answered fast thatd be great.

Question

anonymous · Answer

$$\int\limits_{}^{}\sin^3(2x)\cos^8(2x)dx$$

anonymous · Answer

I have tried so many ways, using a u sub of cos(2x) and changing sin(2x) to 2sinxcosx and even sin^2to 1/2(1-cos(2x))

anonymous · Answer

Anyone?

tkhunny · Answer

What Parts have you tried?  Doesn't seem TOO tricky (albeit rather annoying).

$\int\sin^{3}(2x)\cos^{8}(2x)\;dx = \int\sin^{2}(2x)\;d\left[(-1/18)\cos^{9}(2x)\right]$

$\;\;= -(1/18)\cos^{9}(2x)\sin^{2}(2x) + \int\left[(2/9)\cos^{10}(2x)\right]\cdot \sin(2x)\;dx$

This was quite fruitful, though it may not appear so.  Integration by Parts a second time and you will have it.

tkhunny · Answer

Of course, it may be WAY easier to trade to since for two cosines and make your life easier.

$\sin^{3}(2x)\cos^{8}(2x) = \sin(2x)\left(1-\cos^{2}(2x)\right)\cos^{8}(2x)$

$\;\; = \sin(2x)\left(\cos^{8}(2x) - \cos^{10}(2x)\right)$

Ah, yes.  Now the world is a better place.

anonymous · Answer

@tkhunny  thanks but the answer the book gives is 1/396 cos^9(2 x) (-13+9 cos(4 x)) is this equivelent?

anonymous · Answer

and yea i have got it to that form before, but then coildnt proceed

anonymous · Answer

@tkhunny how does it help find the integral?

anonymous · Answer

still trying to find it anyone can help?

tkhunny · Answer

That last form?  Try $u = \cos(2x)$.  Gotta keep your eyes open...

anonymous · Answer

i have that tried that multiple times i dont think thats gives the right answer...

tkhunny · Answer

??  That makes no sense at all.  If you do it right, you will get the right answer.

Try not to forget that an indefinite integral has that pesky "+C" on the end.  It is NOT optional.  It is what makes $\sin^{2}(x) + C\;and\;\cos^{2}(x) + C$ the SAME result!

anonymous · Answer

Okay so should it be -1/2 u^9/9-u^10/10+c and subing in cos(2x) as u?

anonymous · Answer

u^11/11**

tkhunny · Answer

And, yes, the book's answer is equivalent.  Why not just do it correctly, so you can follow your work, and forget about what the book says.  So far, I have three equivalent expressions.  Shall we start a collection?

anonymous · Answer

Thank you, that is so nice to know

anonymous · Answer

Sure what are the other forms? I want to see if I had got them as well while working on this problem

tkhunny · Answer

The lead directly from the Parts Twice and the Trig + Substitution.  You may explore it.  Obviously, with that $\cos(4x)$, in there, the book used a different trig identity somewhere along the line.

anonymous · Answer

". . . just do it correctly, . . ., and forget about what the book says. "

I think I need to frame this quote and hang it on my wall.