need help with calculus definite integral problme

Question

anonymous · Answer

give me sec to rewrite the problem

superdavesuper · Answer

Nah @YadielG go on - u are doing just fine :)

anonymous · Answer

$$\int\limits_{-2}^{2}\sin(xe^(2x^2)) dx$$

anonymous · Answer

sorry my question was if the 4/x goes on the inside of the sin

superdavesuper · Answer

substituting u=2x^2 into the integral, wat will u get?

anonymous · Answer

du = 4x dx

anonymous · Answer

$$\int\limits_{-2}^{2}\sin((1/4)e^u) du$$

superdavesuper · Answer

crap....the sin( ) will get in the way....

anonymous · Answer

yeah exactly so would it look like i wrote it or would the 4/x be on the outside?

superdavesuper · Answer

no, u cant do the substitution with du=4xdx if the expression is like sin(xe^...)dx as the x is inside the sin(x...)dx :(

anonymous · Answer

hmmm that is what i thought

anonymous · Answer

doing a substitution for the entire expression inside the sin doesn't help either so im kind of stuck on how to approach this

anonymous · Answer

would it be ok to take the natural log of just the inside of the sin( )?

superdavesuper · Answer

u sure it is like this:
$$\int\limits_{-2}^{2}\sin(xe ^{2x ^{2}})dx$$
?
dats pretty complicated...

anonymous · Answer

yes that is the problem

superdavesuper · Answer

no choice but to do the substitution
$$u=xe ^{2x ^{2}}$$
then use product n chain rules to find du....

anonymous · Answer

but how? doesn't make the problem more complicated because there are more terms on the outside of the sin ( )

misty1212 · Answer

lol don't bother

misty1212 · Answer

convince yourself that this function is even

superdavesuper · Answer

just tried n yup - dx cannot be expressed in a nice expression for du :(

perhaps @misty1212 has a better idea!

misty1212 · Answer

so the integral is zero

anonymous · Answer

hmmm that seems more reasonable @misty1212

superdavesuper · Answer

@misty1212 AHHHHHHH DATS not fair!!!!

but thanks!! :)

misty1212 · Answer

$$\int_{-a}^a\text{even function }dx =0$$ that is what it is testing

misty1212 · Answer

you were not supposed to try to find a closed form for the integral

anonymous · Answer

hahaha i feel dumb now but thank you, the problem seemed very out of place with the rest of the assignment until now XD

misty1212 · Answer

math teachers like to try to trick you
don't fall for it

anonymous · Answer

noted and thank you