Question

How do you integrate (sin2pix)^2 with the boundaries being 1/4 on top and 0 at the bottom?

Answer 1

\[\huge \int_0^{1/4} [\sin (2\pi x)]^2dx\]
is that the question?

Answer 2

yup

Answer 3

interesting...no cosines huh

Answer 4

how can you write it like that? Is it a website?

Answer 5

nope...it's a latex code...you can do some basic stuffs by using the "Equation button" below the reply box

Answer 6

oh sorry, there's no 2 in the 2pix

Answer 7

so pi x?

Answer 8

yup

Answer 9

you have learned integration by parts already right?

Answer 10

hm.. I'm not sure what you mean - what's integration by parts?

Answer 11

\[\huge uv - \int vdu\]
or something like that...no?

Answer 12

is it when you use u to replace something? like dy/du x du/dx?

Answer 13

Use this

\[\sin^2 \theta=\frac{1-\cos 2\theta}{2}\]

Answer 14

yup I used that, but I got stuck

Answer 15

let \(\theta\) = \(\pi x\)
first

Answer 16

\[\int\limits_{1/4}^{0}(1-\cos2pix)/2 dx\]

Answer 17

opps I typed it wrong.
Hmm.. ok let theta = \[pix\]

Answer 18

it would be simpler if u let \(\theta=2\pi x\) so \(\theta\) goes from 0 to \(\pi/2\) since \(x\) goes 0 to \(1/4\)

Answer 19

Ok, is \[\pi/8 - 1/4\sin(\pi/2)\] right?
I'll try using \[2 \pi x\] now

Answer 20

@mukushla I tried using \[2 \pi x \] but I ende dup with 1 - 0.5 ((4sin\[\pi /2 \]/pi - sin 0/0)

Answer 21

Sorry that it's so messy

Answer 22

u = 2x
du/2pi = dx

\[\frac{1}{2\pi}\int\limits_{0}^{1/4}\sin^{2}(u)du = \frac{1}{2\pi}\int\limits_{0}^{1/4}\frac{1-\cos(2u)}{2}\]

now you can easily solve this integral

Answer 23

\[\frac{1}{2\pi}\int\limits_{0}^{1/4}\frac{1}{2}du - \int\limits_{0}^{1/4}\frac{\cos(2u)du}{2}\]

Answer 24

note there should be brackets around it but yeah easy to solve now :)

Answer 25

oh that makes sense now. I can solve it now! I'll tell you when I'm done!

Answer 26

yeah there should be brackets around it both integrals should have 1/2pi multiplied into them

Answer 27

ok, I added the brackets and I got
1/16 - 1/8pi is it right?

Answer 28

\[\frac{1}{2\pi}\int\limits_{0}^{1/4}\frac{1}{2}du - \frac{1}{2\pi}\int\limits_{0}^{1/4}\frac{\cos(2u)du}{2}\]

Answer 29

https://www.wolframalpha.com/input/?i=integrate+from+0+to+1%2F4+sin%282pix%29^%282%29

is the answer

Answer 30

oh.. thanks. My answer is really off. Um, do I have to change the boundaries if I'm using "u"?

Answer 31

well you can but I usually prefer to just resub for u after integrating

Answer 32

so replace u with 2pix

Answer 33

after integrating

Answer 34

you will also want to do another substitution on cos(2u)/2 before integrating

Answer 35

ok, I'm working on it = )

Answer 36

Here this will show you how to deal with integrals such as these

Answer 37

this is the source covers most everything done in calculus 1 at my university

Answer 38

wow thanks! It summarises everything!
hmm.. ok I've got 1/8 - 1/4pi now but it still doesn't equal to 1/8

Answer 39

can you scan or take a picture of your work?

Answer 40

ok, I'll try write it up