Question

don't post irrelevant answers...​

Answer 1

Answer:

Using Weierstrass substitution, we get

∫π0dx(5+4cosx)2=∫∞01(5+41−t21+t2)22dt1+t2=2∫∞0t2+1(t2+9)2dt=∫∞−∞t2+1(t2+9)2dt

Now let

f(z)=z2+1(z2+9)2

Then, f has a poles of order 2 at ±3i. We get

resz=3if=limz→3iddzz2+1(z+3i)2=limz→3i(−2(1+z2)(z+3i)3+2z(z+3i)2)=−5i54

Hence, letting γ be the semi circle of radius R>3 centered at the origin and in the upper half-plane, we get that γ encloses the pole z=3i, so

∫γf(z)dz=2πi resz=3if=5π27

Letting R→∞, we see that the integral on the arc-section vanish so that we are left with

∫π0dx(5+4cosx)2=5π27