To determine

To evaluate:

∫-11sinx1+x2 dx is it exists

Answer

0

Explanation

1) Concept:

By using theorem for integration of an odd function

2) Theorem:

Suppose f is continuous on [-a, a] and if f is odd [f-x=-fx], then

∫-aaf(x)dx=0

3) Given:

∫-11sinx1+x2 dx

4) Calculation:

Consider,

∫-11sinx1+x2 dx

Here,

fx=sinx1+x2

Now,

f-x=sin⁡(-x)1+(-x)2

Since sin⁡-x=-sinx

f-x=-sin⁡x1+-x2=-f(x)

Therefore, the function is odd; also it is continuous in thegiven interval, therefore, its integral exists.

From the theorem

∫-11sinx1+x2 dx=0

Conclusion:

Therefore,

∫-11sinx1+x2 dx=0