calculate-0-2pi-cos-5-3cos-d-

Question Number 64649 by mathmax by abdo last updated on 20/Jul/19

c a l c u l a t e \int_{0}^{2 π} \frac{c o s θ}{5 + 3 c o s θ} d θ

Commented by mathmax by abdo last updated on 20/Jul/19

l e t A = \int_{0}^{2 π} \frac{c o s θ}{5 + 3 c o s θ} \Rightarrow A = \frac{1}{3} \int_{0}^{2 π} \frac{3 c o s θ + 5 - 5}{3 c o s θ + 5} d θ

= \frac{2 π}{3} - \frac{5}{3} \int_{0}^{2 π} \frac{d θ}{3 c o s θ + 5} c h a n g e m e n t e^{i θ} = z g i v e

\int_{0}^{2 π} \frac{d θ}{3 c o s θ + 5} = \int_{∣ z ∣= 1} \frac{1}{3 \frac{z + z^{- 1}}{2} + 5} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{2 d z}{i z (3 z + 3 z^{- 1} + 10)} = \int_{∣ z ∣= 1} \frac{- 2 i}{3 z^{2} + 3 + 10 z} d z l e t

W (z) = \frac{- 2 i}{3 z^{2} + 10 z + 3} p o l e s o f W ?

Δ^{^{'}} = 5^{2} - 9 = 16 \Rightarrow z_{1} = \frac{- 5 + 4}{3} = - \frac{1}{3} a n d z_{2} = \frac{- 5 - 4}{3} = - 3 \Rightarrow

W (z) = \frac{- 2 i}{3 (z - z_{1}) (z - z_{2})} (z_{2} i s o u t o f c i r c l e) r e s i d u s t h e o r e m g i v e

\int_{∣ z ∣= 1} W (z) d z = 2 i π R e s (W, z_{1})

R e s (W, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) W (z) = \frac{- 2 i}{3 (z_{1} - z_{2})} = \frac{- 2 i}{3 (- \frac{1}{3} + 3)}

= \frac{- 2 i}{3 . \frac{8}{3}} = - \frac{i}{4} \Rightarrow \int_{∣ z ∣= 1} W (z) d z = 2 i π (- \frac{i}{4}) = \frac{π}{2} \Rightarrow

A = \frac{2 π}{3} - \frac{5}{3} \frac{π}{2} = \frac{2 π}{3} - \frac{5 π}{6} = \frac{4 π - 5 π}{6} \Rightarrow A = - \frac{π}{6} .

Answered by MJS last updated on 20/Jul/19

\underset{0}{\int^{2 π}} \frac{\cos θ}{5 + 3 \cos θ} d θ = 2 \underset{0}{\int^{π}} \frac{\cos θ}{5 + 3 \cos θ} d θ =

[t = \tan \frac{θ}{2} \to d θ = {2 \cos}^{2} \frac{θ}{2} d t]

- 2 \underset{0}{\int^{\infty}} \frac{t^{2} - 1}{(t^{2} + 1) (t^{2} + 4)} d t = \frac{4}{3} \underset{0}{\int^{\infty}} \frac{d t}{t^{2} + 1} - \frac{10}{3} \underset{0}{\int^{\infty}} \frac{d t}{t^{2} + 4} =

= {[\frac{4}{3} \arctan t - \frac{5}{3} \arctan \frac{t}{2}]}_{0}^{\infty} = - \frac{π}{6}

Commented by mathmax by abdo last updated on 20/Jul/19

t h a n k y o u s i r .