Prove the following is true:

Question

Prove that the indefinite integral from 0->pi/2 of cos^10(x) dx = 9/10 times the indefinite integral from 0->pi/2 of cos^8(x) dx

 

This problem showed up on one of the old exams and I fear it may show up on my upcoming one. I'm very lost as how to prove it.

Answer 1

This can be done by deriving the formula for I_n = ∫cosⁿ x dx. This is done by getting the reduction formula via integration by parts.

 

BP: u = sin x ; du = cos x dx ; dv = sin x cos^n-2 x dx ; v = -(cos^n-1 x)/(n-1)

 

I_n = ∫cosⁿ x dx

 

= ∫(1 - sin² x)cos^n-2 x dx

 

= ∫ cos^n-2 x dx - ∫sin² x cos^n-2 x dx

 

= ∫ cos^n-2 x + (sin x cos^n-1 x)/(n-1) - ∫ (cosⁿ x)/(n-1) dx

 

= I_n-2 + (sin x cos^n-1 x)/(n-1) - I_n/(n-1) 

 

Therefore:

 

I_n = [(n-1)/n]I_n-2 + (sin x cos^n-1 x)/n

 

Now, letting A_n be the value of the definite integral, we see that:

 

A_n = [(n-1)/n]A_n-2 + (sin x cos^n-1 x) |₀^π/2 = [(n-1)/n]A_n-2

 

or in other words: A_n/A_n-2 = 1 - 1/n

 

Thus A₁₀/A₈ = 1 - 1/10 = 9/10.