Question

integral

Answer 1

\[\int\limits_{0}^{} \frac{ \cos(2x) dx }{ 9x^2+4 }\]

Answer 2

from 0 to infinity *{infinity is not printed on the integral}*

Answer 3

\[\int\limits_{C}^{}\frac{ e^{i2z} }{ (z+2/3i)(z-2/3i) } dz\]

Answer 4

\[\lim_{z \rightarrow 2/3i}\int\limits_{C}^{}\frac{ e^{i2z} }{ (z+2/3i) }\]

Answer 5

I get \[\frac{ 3 \pi e^{-4/3} }{ 2}\], but the answer is different

Answer 6

http://www.wolframalpha.com/input/?i=cos(2x)%2F(9x%5E2%2B4)+integrate+from+0+to+inf

Answer 7

\[\lim_{z \rightarrow 2/3i}\frac{ e^{i2x} }{ z+(2/3i) }*2*\pi*i\]

Answer 8

^ step 2 in detail

Answer 9

i think your integral is just a tad off

Answer 10

we have this:
\[\Large \int_0^{ + \infty } {} = \frac{1}{2}\int_{ - \infty }^{ + \infty } {} \]

Answer 11

\[\int\limits\limits_0^\infty \frac{e^{2ix}+e^{-2x i}}{ 2} \cdot \frac{1}{(3x+2i)(3x-2i)} dx \\ =\frac{1}{18} \int\limits\limits_0^\infty \frac{e^{2i x}-e^{-2ix}}{(x+\frac{2i}{3})(x-\frac{2i}{3})} dx\]

Answer 12

so if you multiply that result you got by 1/18 you will get wolfram's answer

Answer 13

I agree with @myininaya we have to make this substitution:
\[\Large \cos \left( {2x} \right) = \frac{{{e^{i2x}} + {e^{ - i2x}}}}{2}\]
then we have to go to the complex plane, and we have to apply the theorem of Jordan twice: