Question

Evaluate integral

Answer 1

sin^2+cos^2=1
sin^2=1-cos^2
sin=√(1-cos^2)

Answer 2

\[\Large \int\limits_0^{\pi/8}14\sqrt{1-\cos8x}\;dx\]Recall your `Half-Angle Identity for Sine`:\[\Large \color{#3399AA}{\sin^2\theta\quad=\quad \frac{1}{2}(1-\cos2\theta)}\]

Answer 3

So under the square root we can fiddle around with it a little bit...\[\Large \int\limits\limits_0^{\pi/8}14\sqrt{2\left[\color{#3399AA}{\frac{1}{2}\left(1-\cos8x\right)}\right]}\;dx\]

Answer 4

\[\Large =\int\limits_0^{\pi/8}14\sqrt{2\left[\color{#3399AA}{\sin^24x}\right]}\;dx\]

Answer 5

Now that the subtraction is gone, we can take the root without any troubles!
Any confusion on how I applied that identity?

Answer 6

hm nope its clear, ah so must memorize these identities, sigh.. next is how do u integrate this?

Answer 7

So umm let's take the root of the stuff, giving us,\[\Large 14\sqrt2\int\limits_0^{\pi/8}\sin 4x\;dx\]Understand how to integrate sin4x?
If not, you can always apply a u-sub to make it a little more clear to yourself.

Answer 8

eh isnt the square-root also covers sin4x?

Answer 9

It was a `square` under the `square root`

Answer 10

ah i see okay so its -1/4cos4x, i subbed in the limits i get (7sqrt2)/2 is it correct?

Answer 11

Mmm that's the same thing I'm coming up with, yay team! \c:/
Good job!

Answer 12

okay thanks alot! ^^