Major h.w help needed tri functions derivatives =\

math_student03
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Hey guys having major trouble with these 2 questions, i have a screen shot of the questions and my work , much easier to understand then me typing it in here. thanks soo much for helping !

http://img204.imageshack.us/my.php?image=img050lc1.jpg
 
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You have the right idea, but you just need to be watch out for a few small things.

First of all:
Remember the Chain Rule:
<br /> \frac{d}{dx}(sin(2x)) = 2cos(2x)<br />

Secondly:
Be careful with your minus signs, especially with going from step 2 to step 3.

Finally:
Once you get all that out of the way, go through your trig identities and see if you can't simplify it any (http://www.sosmath.com/trig/Trig5/trig5/trig5.html if you need a good list of them).

- Jason
 
You have an error in the derivitave of sin(2x) and cos(2x). Use the chain rule, you are missing a factor of 2. Continue.
 
ok so basically for my second line i was only missing a 2 for dsin(2x)/dx and dcos(2x)/dx then i can go ahead and simplift using tri propertys? did i mix something up cause i don't see any that are obvious

and secondly, did anyone check out the second question i had on that page, did i do the first and second derivative right?
 
The second one matches what I got.
CC
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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