calculate-0-pi-2-dx-cos-4-x-sin-4-x-

Question Number 34283 by math khazana by abdo last updated on 03/May/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{{c o s}^{4} x + {s i n}^{4} x}

Commented by math khazana by abdo last updated on 08/May/18

l e t p u t I = \int_{0}^{\frac{π}{2}} \frac{d x}{{c o s}^{4} x + {s i n}^{4} x}

I = \int_{0}^{\frac{π}{2}} \frac{d x}{{({c o s}^{2} x + {s i n}^{2} x)}^{2} - 2 {c o s}^{2} x {s i n}^{2} x}

= \int_{0}^{\frac{π}{2}} \frac{d x}{1^{2} - 2 {c o s}^{2} x . {s i n}^{2} x}

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

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

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

= \int_{0}^{\frac{π}{2}} \frac{2 d x}{1 + \frac{1 + c o s (4 x)}{2}} = \int_{0}^{\frac{π}{2}} \frac{4 d x}{3 + \cos (4 x)}

=_{4 x = t} \int_{0}^{2 π} \frac{d t}{3 + c o s t} c h a n g e m e n t 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 ∣= 1} \frac{- 2 i}{6 z + z^{2} + 1} d z

l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n

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

Commented by math khazana by abdo last updated on 08/May/18

Δ^{^{'}} = 3^{2} - 1 = 8 \Rightarrow z_{1} = - 3 + 2 \sqrt{2}

z_{2} = - 3 - 2 \sqrt{2}

∣ z_{1} ∣ - 1 = 2 \sqrt{2} - 3 - 1 = 2 \sqrt{2} - 4 < 0

∣ z_{2} ∣ - 1 = 3 + 2 \sqrt{2} - 1 = 2 + 2 \sqrt{2} > 0 (t o e l i m i n a t e

f r o m r e s i d u s)

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

R e s (φ, z_{1}) = \frac{- 2 i}{(z_{1} - z_{2})} = \frac{- 2 i}{4 \sqrt{2}} = \frac{- i}{2 \sqrt{2}}

\int_{∣ z ∣= 1} φ (z) d z = 2 i π . \frac{- i}{2 \sqrt{2}} = \frac{π}{\sqrt{2}}

I = \frac{π}{\sqrt{2}} .