Integrate
Cos^7(2x)
Explanation would be helpful
Let u=2x du/dx=2 =1/2Incos^7u(int=integral) 1/2Intcos^6u*cosu since cos^2u=1-sin^2u there cos^6u=(cos^2u)^3=(1-sin^2u)^3 there the integral become:1/2Int(1-sin^2u)^3cosudu let v=sinu dv/du=cosu du=dv/cosu substitute it in du and v in integral gives:1/2Int(1-v^2)^3dv expansion of (1-v^2)^3=1-3v^2+3v^4-v^6 so the integral becomes:In 1-In3v^2+In3v^4-Inv^6 integrating each gives:v-v^3+3/5v^5-v^7/7 substitute v=sinu:sinu-sin^3u+3/5sin^5u-1/7sin^7u substitute u=2x:sin2x-sin^3(2x)+3/5sin^5(2x)-1/7sin^7(2x)
wow! what a lot of work.
juat factor out the cosx and expand the (1-sin^2 x)^3
To integrate the function cos^7(2x), we need to use a combination of trigonometric identities and integration techniques. Here's a step-by-step explanation:
Step 1: Apply the power-reducing formula for cosine:
cos(2x) = 1/2 * (1 + cos(4x))
Step 2: Rewrite the original function using the power-reducing formula:
cos^7(2x) = (1/2)^7 * (1 + cos(4x))^7
Step 3: Expand the binomial using the binomial theorem:
cos^7(2x) = (1/128) * (1 + 7cos(4x) + 21cos^2(4x) + 35cos^3(4x) + 35cos^4(4x) + 21cos^5(4x) + 7cos^6(4x))
Step 4: Integrate each term separately:
∫(1/128) * (1 + 7cos(4x) + 21cos^2(4x) + 35cos^3(4x) + 35cos^4(4x) + 21cos^5(4x) + 7cos^6(4x)) dx
The integral of the constant term (1/128) is simple. It becomes (x/128).
For the terms containing cos(4x) raised to an even power, we use the following result:
∫cos^2n(4x) dx = (1/2n) * [(2n-1)x + sin(2n-1)(4x)] + C
For the terms containing cos(4x) raised to an odd power, we use the following result:
∫cos^(2n+1)(4x) dx = (1/2n+1) * sin[2n+1](4x) + C
Step 5: Apply the results for each term:
∫(1/128) * (1 + 7cos(4x) + 21cos^2(4x) + 35cos^3(4x) + 35cos^4(4x) + 21cos^5(4x) + 7cos^6(4x)) dx
= (x/128) + (7/32) * [(1/3)x + sin(3*4x)] + (21/64) * [(1/5)x + sin(5*4x)] + (35/96) * [(1/7)x + sin(7*4x)] + (35/128) * [(1/9)x + sin(9*4x)] + (21/160) * [(1/11)x + sin(11*4x)] + (7/192) * [(1/13)x + sin(13*4x)] + C
where C is the constant of integration.
That's it! You have successfully integrated the function cos^7(2x).