Get the solution to "Computing ∫0π214cos⁡{x}+5, x∈R by substituting cos⁡{x}=eix+e−ix2?"

Harley Ayers

Answered question

2022-03-25

Computing

\int_{0}^{\frac{π}{2}} \frac{1}{4 \cos {x} + 5}, x \in R

by substituting

\cos {x} = \frac{e^{i x} + e^{- i x}}{2}

glikozyd3s68

Beginner2022-03-26Added 16 answers

A fast way is to exploit real methods.

\int_{0}^{\frac{π}{2}} \frac{d x}{5 + 4 \cos x} \overset{x = 2 z}{=} 2 \int_{0}^{\frac{π}{4}} \frac{d z}{1 + 8 \cos^{2} z} \overset{z = \arctan u}{=} 2 \int_{0}^{1} \frac{d u}{9 + u^{2}} =

= \frac{2}{3} \arctan \frac{1}{3}

Alejandra Hanna

Beginner2022-03-27Added 10 answers

This is exactly the Kepler angle substitution:

\sin ψ = \frac{\sqrt{1 - e^{2}} \sin x}{1 + e \cos x}

\cos ψ = \frac{\cos x + e}{1 + e \cos x}

d ψ = \frac{\sqrt{1 - e^{2}} d x}{1 + e \cos x}

For 0

\int_{0}^{\frac{π}{2}} \frac{d x}{5 + 4 \cos x} = \frac{1}{5} \frac{1}{\sqrt{1 - e^{2}}} \int_{0}^{\cos^{- 1} e} d ψ = \frac{\cos^{- 1} e}{5 \sqrt{1 - e^{2}}}

= \frac{\cos^{- 1} (\frac{4}{5})}{3} = \frac{1}{3} \tan^{- 1} (\frac{3}{4}) = \frac{2}{3} \tan^{- 1} (\frac{1}{3})

With

e = \frac{4}{5}

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