calculate-0-2pi-dx-cosx-2sinx-2-

Question Number 127772 by Bird last updated on 02/Jan/21

c a l c u l a t e \int_{0}^{2 π} \frac{d x}{{(c o s x + 2 s i n x)}^{2}}

Answered by mathmax by abdo last updated on 02/Jan/21

I = \int_{0}^{2 π} \frac{dx}{{(cosx + 2 sinx)}^{2}} \Rightarrow I = \int_{0}^{2 π} \frac{dx}{\cos^{2} x + 4 sinx cosx + {4 \sin}^{2} x}

= \int_{0}^{2 π} \frac{dx}{1 + 4 sinxcosx + {3 \sin}^{2} x} = \int_{0}^{2 π} \frac{dx}{1 + 2 \sin (2 x) + 3 \frac{1 - \cos (2 x)}{2}}

= \int_{0}^{2 π} \frac{2 dx}{2 + 4 \sin (2 x) + 3 - 3 \cos (2 x)} = \int_{0}^{2 π} \frac{2 dx}{5 + 4 \sin (2 x) - 3 \cos (2 x)}

\underset{2 x = t}{=} \int_{0}^{4 π} \frac{dt}{5 + 4 sint - 3 cost} = \int_{0}^{2 π} \frac{dt}{4 sint - 3 cost + 5} + \int_{2 π}^{4 π} \frac{dt}{4 sint - 3 cost + 5} (\to t = 2 π + u)

= 2 \int_{0}^{2 π} \frac{dt}{4 sint - 3 cost + 5} =_{e^{it} = z} 2 \int_{∣ z ∣= 1} \frac{dz}{iz (4 \frac{z - z^{- 1}}{2 i} - 3 \frac{z + z^{- 1}}{2} + 5)}

= 2 \int_{∣ z ∣= 1} \frac{dz}{iz (\frac{2}{i} (z - z^{- 1}) - \frac{3}{2} (z + z^{- 1}) + 5)}

= - 2 i \int_{∣ z ∣= 1} \frac{dz}{\frac{2}{i} (z^{2} - 1) - \frac{3}{2} (z^{2} + 1) + 5 z}

= 2 \int_{∣ z ∣= 1} \frac{dz}{2 (z^{2} - 1) - \frac{3}{2} i (z^{2} + 1) + 5 iz} = 4 \int_{∣ z ∣= 1} \frac{dz}{4 (z^{2} - 1) - {3 iz}^{2} - 3 i + 10 iz}

= 4 \int_{∣ z ∣= 1} \frac{dz}{{4 z}^{2} - 4 - {3 iz}^{2} - 3 i + 10 iz}

= 4 \int_{∣ z ∣= 1} \frac{dz}{(4 - 3 i) z^{2} + 10 iz - 4 - 3 i}

poles of w (z) = \frac{1}{(4 - 3 i) z^{2} + 10 iz - 4 - 3 i}

Δ^{^{'}} = {(5 i)}^{2} + (4 - 3 i) (4 + 3 i) = - 25 + 16 + 9 = 0 \to one root

z_{0} = - \frac{b^{^{'}}}{a} = - \frac{5 i}{4 - 3 i} = - \frac{5 i (4 + 3 i)}{25} = - \frac{4 i - 15}{5} \Rightarrow

w (z) = \frac{1}{(4 - 3 i) {(z - z_{0})}^{2}} \Rightarrow \int_{∣ z ∣= 1} w (z) dz = 2 i π Res (w, z_{0})

∣ z_{0} ∣= \frac{1}{5} \sqrt{4^{2} + 15^{2}} > 1 \Rightarrow Res (w, z_{0}) = 0 \Rightarrow I = 0