Integral of complicated trig function.

[dcee27]

Homework Statement

This question has me stumped Please help me get this monster completed.
Compute the definite integral of -2011 to 2011 of (1+sin^2(17t))^2011*sin(sin(-t))dt

Homework Equations

The Attempt at a Solution

 

[spamiam]

I think you can use symmetry to help you. Is the function odd?

 

[atomthick]

Replace 2011 with n. Use integration by parts to get a recurrent relation I(n) = F(I(n-1), I(n-2)...). Use this recurrent relation to find the general value for I(n) and afterthat replace n with 2011.

 

[BruceW]

atomthick has the right idea.
To explain it in my words (to make sure you understand what to do):
When you use the product rule a couple of times, (differentiating the sin to the power bit), hopefully you will see a pattern in how the function changes each time you use product rule. So then you make an equation that gives the total change to the function after n times. then use n=2010, which will effectively get rid of that power 2011, so then you can try using normal method of integration.
(I haven't actually tried this specific problem, but this is generally what I would try doing).

 

[BruceW]

The function looks even.
Actually, dcee27's notation is a bit confusing. In one place he has sin^2(17t) and in another place he has sin(sin(-t)).

 

[atomthick]

spamiam has seen the easy way out, indeed the function is odd.

F(t) = -F(-t) and F(0) = 0.

 

[BruceW]

Why is it odd?
$$(1 + sin^2(17t) )^{2011} \ sin^2(-t)$$
Looks like an even function to me.

 

[atomthick]

I have found that, in general, when a prof gives a rather messy integral over -a to a, the FIRST thing you should do is see if the function you are integrating is odd. If it is, then the value of the integral is 0.

That being said, atomchick has a really cool idea that would be useful when not integrating from -a to a.

Thanks Robert1986 but I'm not an atomCHICK

 

[atomthick]

Why is it odd?
$$(1 + sin^2(17t) )^{2011} \ sin^2(-t)$$
Looks like an even function to me.

I believe the formula is actualy $$(1 + sin(17t)\sin(17t) )^{2011} \ sin(sin(-t))$$ but dcee27 knows better...

 

[BruceW]

Woops, yeah I got sin squared and sin of sin mixed up.

 

[dcee27]

The question is entered correclty. (1+sin^2(17t))^2011 * sin(sin(-t))dt bounded by a = -2011 and b= 2011 o fthe definite intergral.

I still can't compute!

 

[micromass]

The question is entered correclty. (1+sin^2(17t))^2011 * sin(sin(-t))dt bounded by a = -2011 and b= 2011 o fthe definite intergral.

I still can't compute!

Can you show that the function is odd??

 

[dcee27]

Yes, this part b of a previous question that asked to prove that the a integral was an odd function. So I would say yes.

 

[micromass]

So, what is the integral of an odd function?

 

[dcee27]

the first part asked to explain and show that why if f(x0 is an odd function integral a= -a and b = a f(t)dt = 0. That is part a and then part b which is posted asked to compute and explain the integral i posted.

 

[Robert1986]

OK, I want you to do two things:

1) Draw the graph from -5 to 5 of some odd function (in fact, just do a few, sin(x), x^x, x)

2) Look at this graph.

3) What is the area under this curve from -5 to 5 (remember, area under the x-axis is "negative area".)

4) Now use what you have done and see if you can generalise to ANY odd function from -a to a.

 

[BruceW]

So, part a) says that the integral of an odd function from -a to a equals zero.
And you've said that the function you were given for part b) is an odd function.
So simply use the fact from part a) to find out the answer to b) (its simpler than you think).

 

[Robert1986]

OK, I want you to do two things:

1) Draw the graph from -5 to 5 of some odd function (in fact, just do a few, sin(x), x^x, x)

2) Look at this graph.

3) What is the area under this curve from -5 to 5 (remember, area under the x-axis is "negative area".)

4) Now use what you have done and see if you can generalise to ANY odd function from -a to a.

There are three types of people in the world: those who can count and those who can't.

Its good to see I belong in the first. :)

 

[Bohrok]

OK, I want you to do two things:

1) Draw the graph from -5 to 5 of some odd function (in fact, just do a few, sin(x), x^x, x)

Erm, x^x isn't an odd function either...

 

[Robert1986]

I'm not sure what I mean by x^x; I'm guessing I meant x^2 or something.

 

[lucifer130621]

I believe the formula is actualy $$(1 + sin(17t)\sin(17t) )^{2011} \ sin(sin(-t))$$ but dcee27 knows better...

Thanks atomthik