Question

Please help guys :)

Answer 1

the box part
http://screencast.com/t/pE8Bnkcu2

Answer 2

i got \[2 \cos (\theta-\frac{\pi}{3})\]

Answer 3

i always forget this and so i refer to this to refresh my feeble memory

Answer 4

hehe...

Answer 5

i think it is \(2\cos(x-\frac{\pi}{6})\)

Answer 6

it is easy to get the \(a\) and \(b\) backwards

Answer 7

its \[\frac{\pi}{3}\] im pretty sure

Answer 8

tan theta = sqrt(3)/1 = sqrt(3)
with theta in the 1st quadrant
yes, it pi/3

Answer 9

lol i told you it was easy to get it backwards....

Answer 10

you are right, i am wrong
sorry

Answer 11

@Loser66 it is not always arctan because you might be in quadrant 2 or 3

Answer 12

my problem is the box? i hope u all would help me with that

Answer 13

but now you are basically done right? you have
\[\frac{1}{2}\int \sec^2(x-\frac{\pi}{3})dx\]

Answer 14

why is it sec?

Answer 15

it doesnt appear that changing one to the other amounts to anything other than an identity, so no "u subs"

Answer 16

because \(\frac{1}{\cos^2(x)}=\sec^2(x)\)

Answer 17

and you know an anti derivative of secant squared in your head

Answer 18

wow..did not notice that

Answer 19

probably forgot about that square

Answer 20

\[\frac1{R}\int\frac{1}{cos^2(\theta-\alpha)}~d\theta\]
let u = theta-alpha
du = dtheta

Answer 21

alright then

Answer 22

just hint :)
derivative of tan(x) is sec^2 x
derivative of tan(x - k) is sec^2 (x - k) (with k is constant)

Answer 23

so its \[\frac{1}{2} \times \tan (\theta-\frac{\pi}{3})\]

with the limits @satellite73

Answer 24

correct!

Answer 25

im nt getting the answer though

Answer 26

1/2 tan(pi/2 - pi/3) - 1/2 tan(0 - pi/3)
= 1/2 1/sqrt(3) - 1/2 tan(-pi/3)
= 1/2sqrt(3) + 1/2 tan(pi/3)
= 1/2sqrt(3) + 1/2sqrt(3)
= 2/2sqrt(3)
= 1/sqrt(3)

Answer 27

thanks

Answer 28

yw

Answer 29

\[\cos \theta+\sqrt{3}\sin \theta =2\left( \frac{ 1 }{ 2}\cos \theta+\frac{ \sqrt{3} }{ 2 }\sin \theta \right)\]
\[=2\left( \cos \frac{ \pi }{ 3 } \cos \theta+\sin \frac{ \pi }{ 3 }\sin \theta \right)=2\cos \left( \frac{ \pi }{ 3}-\theta \right)\]
\[=2\cos \left( \theta-\frac{ \pi }{ 3} \right), \cos \left( -\theta \right)=\cos \theta \]