Question

Evaluate the following integral, if it exists..

Answer 1

\[\int\limits_{-1}^{1} \frac{sinx}{1 + x^2}dx\]

Answer 2

So i can do \[\int\limits_{-1}^{1}sinx * \frac{1}{1+x^2}dx\] and replace the second term with tan^-1(x) ??

Answer 3

like this \[\int\limits_{-1}^{1}sinx \space \tan^{-1}x\]???

Answer 4

i dont believe that is correct. Not that i really have any suggestions unfortunately, but I dont believe you can just switch it out like that.

Answer 5

You could do a u-substitution, and let:\[u=\tan^{-1}(x)\Longrightarrow du=\frac{1}{1+x^2}dx\]but while this gets rid of the fraction, it makes the other part more complicated.

Answer 6

The integral would become:\[\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin(\tan(u))du\]

Answer 7

And then i would have to find anti derivative of that mess and then do back substitution??

Answer 8

right. You might want to type this into wolframalpha just to make sure its solvable before you spend hours trying to solve it. If wolfram gives you an answer in terms of unsolvable integrals, its not possible to solve by hand then.

Answer 9

It says it should be 0

Answer 10

http://www.wolframalpha.com/input/?i=integrate+%28sinx%29%2F%281%2Bx^2%29+from+x%3D-1+to+1

Answer 11

Is there a way to make it show the steps on how it got that answer??

Answer 12

LOLOL omg.

Answer 13

Just prove that:\[\frac{\sin x}{1+x^2}\]is an odd function. and you are done.

Answer 14

Its funny how a picture helps that much. Let:\[f(x)=\frac{\sin x}{1+x^2}\]Show that:\[f(-x)=-f(x)\]so we can claim that f(x) is an odd function for all x. For odd functions, it is always true that:\[\int\limits_{-a}^{a}f(x)dx=0\]since the area to the left of the y axis is equal in magnitude to the area on the right side, but of opposite sign.

Answer 15

Oh ok