find-the-value-of-0-pi-dx-2cos-2-x-sin-2-x-

Question Number 28815 by abdo imad last updated on 30/Jan/18

f i n d t h e v a l u e o f \int_{0}^{π} \frac{d x}{2 {c o s}^{2} x + {s i n}^{2} x} .

Answered by mrW2 last updated on 31/Jan/18

\int_{0}^{π} \frac{d x}{2 {c o s}^{2} x + {s i n}^{2} x}

= \int_{0}^{π} \frac{d x}{1 + {c o s}^{2} x}

= 2 \int_{0}^{π} \frac{d x}{3 + 2 {c o s}^{2} x - 1}

= 2 \int_{0}^{π} \frac{d x}{3 + \cos 2 x}

= 2 \times 2 {[\frac{\tan^{- 1} (\frac{\tan x}{\sqrt{2}})}{2 \sqrt{2}}]}_{0}^{π / 2}

= \frac{π}{\sqrt{2}}

Commented by abdo imad last updated on 31/Jan/18

a n o t h e r m e t o d b y r e d i d u s t h e o r e m w e h a v e

I = \int_{0}^{π} \frac{d x}{1 + {c o s}^{2} x} = \int_{0}^{π} \frac{d x}{1 + \frac{1 + c o s (2 x)}{2}} = \int_{0}^{π} \frac{2 d x}{3 + c o s (2 x)}

t h e c h . 2 x = t g i v e I = \int_{0}^{2 π} \frac{d t}{3 + c o s t} a n d t h e c h . e^{i t} = z g i v e

I = \int_{∣ z ∣= 1} \frac{1}{3 + \frac{z + z^{- 1}}{2}} \frac{d z}{i z} = \int_{∣ z ∣= 1} \frac{2 d z}{i z (6 + z + z^{- 1})}

= \int_{∣ z ∣} \frac{- 2 i}{6 z + z^{2} + 1} = \int_{∣ z ∣= 1} \frac{- 2 i}{z^{2} + 6 z + 1} d z l e t p u t

f (z) = \frac{- 2 i}{z^{2} + 6 z + 1}, p o l e s o f f ?

z^{2} + 6 z + 1 = 0 \Rightarrow Δ = 36 - 4 = 32 \Rightarrow z_{1} = \frac{- 6 + 4 \sqrt{2}}{2} = - 3 + 2 \sqrt{2}

z_{2} = - 3 - 2 \sqrt{2} w e h a v e ∣ z_{1} ∣ - 1 = 2 \sqrt{2} - 3 - 1 = 2 (\sqrt{2} - 2) < 0

a n d ∣ z_{2} ∣ - 1 = 3 + 2 \sqrt{2} - 1 = 2 + 2 \sqrt{2} > 1 (t o e l i m i n a t e b e c a u s e

o u t o f c i r c l e)

\int_{∣ z ∣= 1} f (z) d x = 2 i π R e (f, z_{1}) b u t f (z) = \frac{- 2 i}{(z - z_{1}) (z - z_{2})}

R e s (f, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) f (z) = \frac{- 2 i}{z_{1} - z_{2}} = \frac{- 2 i}{4 \sqrt{2}}

\int_{∣ z ∣= 1} f (z) d x = 2 i π . \frac{- i}{2 \sqrt{2}} = \frac{π}{\sqrt{2}} s o I = \frac{π}{\sqrt{2}} .