integral[(sin(y)*cos(y))/(9+(cos(y))^4)]

i know the answer, but i am stuck on how to find it accurately.

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- Mar 1st 2007, 03:15 PMthedogeneed help with an integral
integral[(sin(y)*cos(y))/(9+(cos(y))^4)]

i know the answer, but i am stuck on how to find it accurately. - Mar 1st 2007, 03:44 PMJhevon
Here's one way to do it.

int{sinycosy/(9 + cos^4(y)}dy

= int{sinycosy/[9 + (cos^2(y))^2]}dy

Now we proceed by substitution:

Let u = cos^2y

=> du = -2cosysiny dy

=> (-1/2)du = cosysiny dy

So our integral becomes

(-1/2)*int{1/(9 + u^2)}du

= (-1/2)*int{1/(3^2 + u^2)}du

= (-1/2)*int{(1/9)*(1/[1 + (u^2)/9])}du

= (-1/2)(1/9)*int{1/(1 + (u/3)^2)}du

Substitution again:

Let v = u/3

=> dv = 1/3 du

=> 3dv = du

So our new integral is:

(-1/18)(3)int{1/(1 + v^2)}dv

= (-1/6)int{1/(1 + v^2)}dv

= (-1/6)arctan(v) + C

= (-1/6)arctan(cos^2(y)/3) + C

This method is correct, but there are other ways to do it that will end up with answers that look different. If you type this function in the "integrator" for instance, the answer looks different from what we have here. But any difference in the answer can be resolved by algebraic manipulation, if not, the difference is absorbed in the constant. - Mar 1st 2007, 04:19 PMJhevon
- Mar 1st 2007, 04:26 PMthedoge
yea. i got it right. i was just confused about how to approach the middle section to reach the point where we got the integral to int(1/(1+x^2))=arctan... :) you cleared up for me, so i said thanks.