MHB Evaluate Definite Integral Challenge

AI Thread Summary
The discussion revolves around evaluating the definite integral $$\int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx$$. Participants express their thoughts on using calculators for solving the problem, with one member humorously acknowledging reliance on a calculator for the solution. There is a sense of camaraderie among contributors, with appreciation for prompt responses. The conversation highlights the balance between manual calculations and technological assistance in solving mathematical challenges. Overall, the thread emphasizes the collaborative nature of tackling complex integrals.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Evaluate $$\int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx$$
 
Mathematics news on Phys.org
I'm sure everyone here would consider it cheating, but my handy-dandy HP 50g yields
$$ \int_{0}^{ \pi} \frac{ \cos(4x)- \cos(4 \alpha)}{ \cos(x)- \cos( \alpha)} \, dx=8 \pi \cos^{3}( \alpha)-4 \pi \cos( \alpha)=4 \pi \cos( \alpha)[2 \cos^{2}( \alpha)-1]=4 \pi \cos( \alpha) \cos(2 \alpha).$$
 
My solution:

If we apply a double-angle identity for cosine on the numerator of the integrand, we find:

$$\cos(4x)-\cos(4\alpha)=\left(2\cos^2(2x)-1 \right)-\left(2\cos^2(\alpha)-1 \right)=$$

$$2\left(\cos(2x)+\cos(2\alpha) \right)\left(\cos(2x)-\cos(2\alpha) \right)$$

Applying a double-angle identity for cosine again, we have:

$$4\left(\cos(2x)+\cos(2\alpha) \right)\left(\cos(x)+\cos(\alpha) \right)\left(\cos(x)-\cos(\alpha) \right)$$

Hence, the definite integral may now be written:

$$I=4\int_0^{\pi}\left(\cos(2x)+\cos(2\alpha) \right)\left(\cos(x)+\cos(\alpha) \right)\,dx$$

Using the property $$\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$$ we also have:

$$I=4\int_0^{\pi}\left(\cos(2x)+\cos(2\alpha) \right)\left(-\cos(x)+\cos(\alpha) \right)\,dx$$

Adding the two expressions for $I$, we obtain:

$$2I=8\cos(\alpha)\int_0^{\pi}\cos(2x)+\cos(2\alpha)\,dx$$

Hence:

$$I=4\pi\cos(\alpha)\cos(2\alpha)$$

Adrian, glad to see we obtained the same result! :D
 
Thanks for participating, Ackbach and MarkFL and I think both of you deserve a pat on the back for this prompt reply...though Ackbach did depend on a little gizmo to obtain the answer, hehehe...
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top