# Integral question

1. Jan 16, 2008

### transgalactic

x^2 * [1+(sinx)^2007]dx

the integral is from 1 to -1

i tried to solve it by parts but its too complicated

2. Jan 16, 2008

### Tom Mattson

Staff Emeritus
Have you learned anything about integrating even and odd functions?

3. Jan 16, 2008

### rocomath

Can you tell me a little more? Thanks :-]

4. Jan 16, 2008

### Mathdope

Even functions satisfy f(-x) = f(x). Odd functions satisfy f(-x) = -f(x). Note the interval over which you are integrating makes this rather useful.

5. Jan 17, 2008

### transgalactic

thats not an odd functon my function is composed of two different function
which are different from one another
f(-x) = -f(x)

cosx is an odd function ,
but here i am not sure
if i knew its odd then we need to split it into two different intervals
and still find the integral

i tried to solve it by parts but thats not working

6. Jan 17, 2008

### TheoMcCloskey

Indeed - but it is a linear combination of two functions - one even and one odd.

7. Jan 17, 2008

### Gib Z

Think of the integrand as one entire function, not the composition of separate ones. Check if its even or odd. If you haven't run into integrating odd and even functions before, you can interpret it algebraically or geometrically.

Algebraically, split the integral over the intervals -1 to 0 and 0 to 1. Then use the odd function property.

Geometrically, do you see any symmetry when you graph the functions?

8. Jan 17, 2008

### HallsofIvy

Staff Emeritus
By symmetry, if f(x) is an even function,
$$\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx$$
If f(x) is an odd function,
$$\int_{-a}^a f(x)dx= 0$$

Write your integrand as x2+ x2 sin2007(x).

What is (-x)2? What is (-x)2 sin2007(-x)?

9. Jan 17, 2008

### transgalactic

my problem is to solve the actual function to find the previos function
on which we made the derivative

by parts is not working

10. Jan 17, 2008

### silver-rose

Expand the product out, and take the integral of each of the terms, using the idea of odd and even functions.

Why would u use by-parts in the first place? What are you trying to reduce to zero power? If you're trying to eventually take x^2 down to x^0, I don't see how you can in turn, integrate sin(x)^2007.

11. Jan 17, 2008

### transgalactic

i read the post and in the end i am still required to solve
x^2 * (sinx)^2007

and i dont have any clue how to do it

12. Jan 17, 2008

### Gib Z

$$f(x) = x^2 \cdot \sin^{2007} (x)$$

$$f(-x) = (-x)^2 \cdot \sin^{2007} (-x) = x^2 \cdot \left( - \sin^{2007} x\right)= - x^2 \cdot \sin^{2007} (x) = - f(x)$$

Repeating;
$$f(-x) = - f(x)$$.

$$\int^a_{-a} f(x) dx = \int^0_{-a} f(x) dx + \int^a_0 f(x) dx$$.

In the first integral, letting x = -u, dx = - du. The bounds change from 0 to 0, and -a to a.

$$\int^0_{a} - f(-u) du = \int^0_a f(x) dx = - \int_0^a f(x) dx$$.

I probably shouldn't have given you the whole answer, but I trust that you will read over the solution several times so you understand it perfectly, or else I just did a misdeed.

13. Jan 17, 2008

### transgalactic

but you didnt finish it
the main problem is to solve f(x)dx
from a to 0
or the other one

iknow of the proccess of spliting the interval
the main problem for me is to solve this integral f(x)dx

i tried to solve it in part ,in substitution
nothing works

14. Jan 17, 2008

### Gib Z

Do you just want an anti derivative or do you want the smart way???

I practically did it for you!!

$$\int^a_{-a} f(x) = \int^a_0 f(x) dx - \int^a_0 f(x) dx$$.What is the right hand side equal to?

15. Jan 17, 2008

### transgalactic

it equals to zero
wow
you switched the intervals

16. Jan 17, 2008

### Gib Z

=] So you fine with this now? We can go over other examples if you need to.

17. Jan 17, 2008

### transgalactic

thanks

18. Jan 17, 2008

### Defennder

I'm wondering, does anyone know how to evaluate the function as an indefinite integral? Any hints?

19. Jan 17, 2008

### Gib Z

The anti derivative will be ugly, if it exists in elementary form.

20. Jan 23, 2008

### transgalactic

why its zero???

because if we draw the function we get two areas which are
opposite one to another

bu still these areas exists and we need to count them
they both equal to each other
so we need to find the area of one of them
and multiply the result by 2

???

why its zero
zero is a mistake
we need to find the area of one of them
and multiply the result by 2