Expert Answer to "Prove that∫0πcos⁡xcos⁡4x(2−cos⁡x)2dx=π9(2160−12473)"

calcolare45pj

Answered question

2022-03-31

Prove that

\int_{0}^{π} \frac{\cos x \cos 4 x}{{(2 - \cos x)}^{2}} d x = \frac{π}{9} (2160 - 1247 \sqrt{3})

anghoelv1lw

Beginner2022-04-01Added 19 answers

We have

\int_{0}^{π} \frac{\cos m x}{a^{2} - 2 a b \cos x + b^{2}} d x = \frac{π}{a^{2} - b^{2}} {(\frac{b}{a})}^{m}

for

| b | < a

(1)
Now, let

p = a^{2} + b^{2}

and

q = 2 a b

, then

p + q = \sqrt{p + q}

and

\sqrt{p - q}

Therefore

2 a = \sqrt{p + q} + \sqrt{p - q}

2 b = \sqrt{p + q} - \sqrt{p - q}

a^{2} - b^{2} = \sqrt{p^{2} - q^{2}}

(2)

\frac{b}{a} = \frac{p - \sqrt{p^{2} - q^{2}}}{q}

(3)
then plugging in (2) and (3) to (1) we prove our proposition,
Set

m = 4

and

p = 2

then differentiate the proposition w.r.t. q and take the limit for

q \to 1

, we obtain

lim_{q \to 1} \int_{0}^{π} d_{q} (\frac{\cos 4 x}{2 - q \cos x}) d x = lim_{q \to 1} d_{q} (\frac{π}{\sqrt{4 - q^{2}}} {(\frac{2 - \sqrt{4 - q^{2}}}{q})}^{4})

\int_{0}^{π} \frac{\cos x \cos 4 x}{{(2 - \cos x)}^{2}} d x = \frac{π}{9} (2160 - 1247 \sqrt{3})

The last step is confirmed by Wolfram Alpha.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?