Question

Really stuck with this one. See attachment.

Answer 1

I got y = 0.013 but I know it's wrong.

Answer 2

According to standard table of integrals,

cos kθ = 1/k sin kθ +c

But what does this mean in relation to solving this question?

Answer 3

What are you stuck on?

Answer 4

\[\int\limits_{0}^{\frac{\pi}{4}}\cos2\theta ~d\theta\] integrate and evaluate at the limits.

Answer 5

How do I integrate it?

Answer 6

As you pointed out, the integral of cos kθ = 1/k sin kθ +c , so you must plug in you 'k' and then plug in your upper limit and subtract the equation containing your lower limit.

Answer 7

What would my 'k' be in this question?

Answer 8

Aha k = 2

Answer 9

Yes.

Answer 10

so what do I do with this: \[\cos2\theta = \frac{ 1 }{ 2 }\sin2\theta + c\]

Answer 11

You do not need the 'c' right now, but you take that equation and plug in your upper limit, then you take that equation and plug in your lower limit, finally you subtract the "low limit" results from the "up limit" results.

Answer 12

\[[ \cos2\theta = \frac{ 1 }{ 2 }\sin2\theta ]^{\frac{ \pi }{ 4 }} - [\cos2\theta = \frac{ 1 }{ 2 }\sin2\theta ]^{0}\]

Answer 13

Technically, yes. However...

Answer 14

First your limit IS theta, second, you don't need your original equation in there. \[[\frac{ 1 }{ 2 }\sin{(2*\frac{ \pi }{ 4 })}] - [\frac{ 1 }{ 2 }\sin{(2*0)}]\]

Answer 15

Since you are integrating over a definite period you no longer have to worry about 'c'.

Answer 16

Oh I see!

Answer 17

Gimme a minute to work this out.

Answer 18

I got 0.14 (2dp) as my answer but that's wrong, as its the same answer I got last time when I tried to work this question out.

Answer 19

Strange because I got 1/2

Answer 20

Oh, you didn't calculate using radians.

Answer 21

So by putting the calculator in radian mode it'll work?