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=-sinx1+-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