Question

How do I solve
16 int_{0}^{pi/6} sqrt{1-sin ^2 theta} cos theta d theta ?

Answer 1

\[16 \int\limits_{0}^{\pi/6} \sqrt{1-\sin ^2 \theta} \cos \theta d \theta\]

Answer 2

@.Sam. @UnkleRhaukus

Answer 3

I was going to let sin theta = u, but then I got stuck.

Answer 4

change the radical to cos theta
you get cos^2 theta d theta

now use the identity cos^2(x) = (1+ cos(2x))/2

Answer 5

ohhh...

Answer 6

cos^2 x + sin^2 x = 1

Answer 7

Just a sec.

Answer 8

okay, now I have \[8 \int\limits_{0}^{\pi/6} 1+ \cos^2 \theta d \theta \]

Answer 9

yes

Answer 10

Wait, that should be cos(2 theta)

Answer 11

yes 2 theta

Answer 12

Now I have \[8[\theta - (1/2) \sin(2 \theta) from 0 \to \pi/6\]

Answer 13

ok

Answer 14

though I think that should be + sin(2x)

Answer 15

as a check: d (+sin 2x) = cos 2x 2 dx