Question

Simple question about intergation

Answer 1

Part 2

Answer 2

The question is of 2 marks so not a lot of working is required for the second part

Answer 3

My approach:
1) First, work out the bounds of A
2) u = sin2x => du = 2cos2x dx
3) (ii) would very much make use the answer for (I)

Answer 4

I know that I just dont know how to use it
@Kira_Yamato

Answer 5

Ok we'll start off with step (1):
We can easily say that either sin x = 0 or cos x = 0 (0 < x < pi/2)
x = 0 or pi/2 (i.e. the region A is bound by x-coordinates 0 and pi/2)

Acceptable to this point?

Answer 6

Yes

Answer 7

Next?

Answer 8

u = sin 2x => x = 0.5 arcsin u => \[dx = \frac{du}{1 - \sqrt(1-x^2)}\]

Answer 9

Can you just use the answers from part i

Answer 10

Ok that's possible as well:
\[u^2 (1-u^2) = \frac{10}{3}\]
\[u^4 - u^2 + \frac{10}{3} = 0\]
You can solve it as a quadratic equation in u^2

Answer 11

Its a two mark question

Answer 12

Are we done with the first part? What did \(A\) come out to be?

Answer 13

1/24

Answer 14

Yes, that's correct. You should notice that \(|\sin^3 2x \cos^3 2x|\) is PERIODIC with period \(x = \pi/4\). We found that\[\int_{0}^{\pi/4}\sin^3 2x \cos^3 2xdx = \int_{0}^{\pi/4} |\sin^3 2x \cos^3 2x| dx = A \]In that case\[40A = \int_{0}^{\color{red}{40}\cdot \pi/4} |\sin^3 2x \cos^3 2x| dx\]

Answer 15

Wait a minute how are you so sure that multiplying the area by 40 will also multiply the limit by 40 too

Answer 16

Well, it's periodic, isn't it? Look at it like this:\[\int_{0}^{\pi/4} |\sin^3 2x \cos^3 2x| dx =\int_{n\pi/4}^{(n+1) \pi/4} |\sin^3 2x \cos^3 2x| dx\]

Answer 17

What do you mean by periodic? That the area after each pi/4 is same

Answer 18

The best way, though, is the graph.