nice-calculus-evaluate-0-2pi-1-1-cos-4-x-dx-

Question Number 137155 by mnjuly1970 last updated on 30/Mar/21

\dots \dots n i c e c a l c u l u s \dots \dots

e v a l u a t e ::

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

Answered by Ar Brandon last updated on 30/Mar/21

ϕ = \int_{0}^{2 π} \frac{dx}{1 + \cos^{4} x} = 4 \int_{0}^{\frac{π}{2}} \frac{\sec^{4} x}{\sec^{4} x + 1} dx, t = tanx

= 4 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} + 1)}^{2} + 1} dt = 4 \int_{0}^{\infty} \frac{t^{2} + 1}{t^{4} + {2 t}^{2} + 2} dt

= 2 \int_{0}^{\infty} \frac{(t^{2} + \sqrt{2}) + (t^{2} - \sqrt{2})}{t^{4} + {2 t}^{2} + 2} dt + \frac{2}{\sqrt{2}} \int_{0}^{\infty} \frac{(t^{2} + \sqrt{2}) - (t^{2} - \sqrt{2})}{t^{4} + {2 t}^{2} + 2} dt

= 2 \int_{0}^{\infty} {\frac{(1 + \frac{\sqrt{2}}{t^{2}}) + (1 - \frac{\sqrt{2}}{t^{2}})}{t^{2} + 2 + \frac{2}{t^{2}}}} dt + \frac{2}{\sqrt{2}} \int_{0}^{\infty} {\frac{(1 + \frac{\sqrt{2}}{t^{2}}) - (1 - \frac{\sqrt{2}}{t^{2}})}{t^{2} + 2 + \frac{2}{t^{2}}}} dt

= 2 \int_{0}^{\infty} {\frac{1 + \frac{\sqrt{2}}{t^{2}}}{{(t - \frac{\sqrt{2}}{t})}^{2} + 2 \sqrt{2} + 2} + \frac{1 - \frac{\sqrt{2}}{t^{2}}}{{(t + \frac{\sqrt{2}}{t})}^{2} + 2 - 2 \sqrt{2}}} dt

+ \frac{2}{\sqrt{2}} \int_{0}^{\infty} {\frac{1 + \frac{\sqrt{2}}{t^{2}}}{{(t - \frac{\sqrt{2}}{t})}^{2} + 2 \sqrt{2} + 2} - \frac{1 - \frac{\sqrt{2}}{t^{2}}}{{(t + \frac{\sqrt{2}}{t})}^{2} + 2 - 2 \sqrt{2}}} dt

= 2 \int_{- \infty}^{+ \infty} \frac{du}{u^{2} + 2 \sqrt{2} + 2} + \frac{2}{\sqrt{2}} \int_{- \infty}^{+ \infty} \frac{dv}{v^{2} + 2 \sqrt{2} + 2}

= {[\frac{2}{\sqrt{2 \sqrt{2} + 2}} \tan^{- 1} (\frac{u}{\sqrt{2 \sqrt{2} + 2}}) + \frac{\sqrt{2}}{\sqrt{2 \sqrt{2} + 2}} \tan^{- 1} (\frac{v}{\sqrt{2 \sqrt{2} + 2}})]}_{- \infty}^{+ \infty}

= \frac{2 π}{\sqrt{2 \sqrt{2} + 2}} + \frac{\sqrt{2} π}{\sqrt{2 \sqrt{2} + 2}}

Commented by Ar Brandon last updated on 30/Mar/21

Not 100 % sure \dots Please check

Commented by Dwaipayan Shikari last updated on 30/Mar/21

\frac{2 + \sqrt{2}}{\sqrt{2 \sqrt{2} + 2}} π = \sqrt{1 + \sqrt{2}} π

Commented by mnjuly1970 last updated on 30/Mar/21

t h a n k s a l o t m r b r a n d o n . .

Commented by mnjuly1970 last updated on 30/Mar/21

g r a t e f u l m r p a y a n \dots

Commented by Ar Brandon last updated on 30/Mar/21

Cool

Answered by mathmax by abdo last updated on 31/Mar/21

Φ = \int_{0}^{2 π} \frac{dx}{1 + \cos^{4} x} \Rightarrow Φ = \int_{0}^{2 π} \frac{dx}{1 + {(\frac{1 + \cos (2 x)}{2})}^{2}}

= \int_{0}^{2 π} \frac{4 dx}{4 + (1 + 2 \cos (2 x) + \cos^{2} (2 x))} = 4 \int_{0}^{2 π} \frac{dx}{5 + 2 \cos (2 x) + \frac{1 + \cos (4 x)}{2}}

= 8 \int_{0}^{2 π} \frac{dx}{10 + 4 \cos (2 x) + 1 + \cos (4 x)} = 8 \int_{0}^{2 π} \frac{dx}{11 + 4 \cos (2 x) + \cos (4 x)}

=_{2 x = t} 8 \int_{0}^{4 π} \frac{dt}{2 (11 + 4 cost + \cos (2 t))} = 4 \int_{0}^{4 π} \frac{dt}{11 + 4 cost + {2 \cos}^{2} t - 1}

= 4 \int_{0}^{4 π} \frac{dt}{{2 \cos}^{2} t + 4 cost + 10} = 2 \int_{0}^{4 π} \frac{dt}{\cos^{2} t + 2 cost + 5}

z^{2} + 2 z + 5 = 0 \to Δ^{^{'}} = 1 - 5 = - 4 \Rightarrow z_{1} = - 1 + 2 i and z_{2} = - 1 - 2 i \Rightarrow

\cos^{2} t + 2 cost + 5 = (cost - z_{1}) (cost - z_{2}) \Rightarrow

\frac{1}{\cos^{2} t + 2 cost + 5} = \frac{1}{(cost - z_{1}) (cost - z_{2})}

= (\frac{1}{cost - z_{1}} - \frac{1}{cost - z_{2}}) \times \frac{1}{4 i} \Rightarrow

Φ = \frac{1}{2 i} \int_{0}^{4 π} \frac{dt}{cost - z_{1}} - \frac{1}{2 i} \int_{0}^{4 π} \frac{dt}{cost - z_{2}}

= - i \int_{0}^{2 π} \frac{dt}{cost - z_{1}} + i \int_{0}^{2 π} \frac{dt}{cost - z_{2}} = - iH + iK

H =_{z = e^{it}} \int_{∣ z ∣= 1} \frac{dz}{iz (\frac{z + z^{- 1}}{2} + 1 - 2 i)}

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

φ (z) = \frac{- 2 i}{z^{2} + (2 - 4 i) z + 1} poles ?

Δ^{^{'}} = {(1 - 2 i)}^{2} - 1 = 1 - 4 i - 4 - 1 = - 4 (1 + i) = - 4 \sqrt{2} e^{\frac{i π}{4}} \Rightarrow

\sqrt{Δ^{,}} = 2 i (^{4} \sqrt{2}) e^{\frac{i π}{8}} \Rightarrow z_{1} = - 1 + 2 i + 2 i (^{4} \sqrt{2}) e^{\frac{i π}{8}}

= - 1 + 2 i + 2 i (^{4} \sqrt{2}) (\frac{\sqrt{2 + \sqrt{2}}}{2} + i \frac{\sqrt{2 - \sqrt{2}}}{2})

= - 1 + 2 i + 2 i (^{4} \sqrt{2}) \frac{\sqrt{2 + \sqrt{2}}}{2} - (^{4} \sqrt{2}) \sqrt{2 - \sqrt{2}}

= - 1 + 2 i + i (^{4} \sqrt{2}) \sqrt{2 + \sqrt{2}} - (^{4} \sqrt{2}) \sqrt{2 - \sqrt{2}}

\Rightarrow∣ z_{1} ∣= \sqrt{{(1 + (^{4} \sqrt{2}) \sqrt{2 - \sqrt{2}})}^{2} + \sqrt{2} (2 + \sqrt{2})} \dots . be continued \dots .