I got a email twice saying I have a reply on this thread? (by bjalongi and MathGuru). But nothing here? Is there a problem with MHF site? or did some1 delete their post?
Hello, dadon!
∫ sin^2(x) cos^6(x) dx
Even-powered sines and cosines are the worst case.
It requires both double-angle identities.
. . . sin²θ .= .½(1 - cos2θ) . . . . cos²θ .= .½(1 + cos2θ)
The first factor is: .sin²(x) .= .½(1 - cos2x)
The second factor is: .cos^6(x) .= .(cos²x)³ .= .[½(1 + cos2x)]³
. . . = .(1/8) (1 + 3cos2x + 3cos²2x + cos³2x)
Multiply: . ½(1 + cos2x)·(1/8)(1 + 3cos2x + 3cos²2x + cos³2x)
. . . . . . = .(1/16) [1 + 2cos(2x) - 2cos³(2x) - cos^4(2x)] [1]
The last term is: . cos^4(2x) .= .[cos²2x]² .= .[½(1 + cos4x)]²
. . . . . . = .¼[1 + 2cos4x + cos²4x] .= .¼[1 + 2cos4x + ½(1 + cos8x)]
. . . . . . = .(3/8) + (1/2)cos4x + (1/8)cos8x
Substitute into [1]: .(1/16) [1 + 2cos2x - 2cos³2x - (3/8) - (1/2)cos4x - (1/8)cos8x]
. . . . . = .(1/16) [(5/8 ) + 2cos2x - 2cos³2x - (1/2)cos4x - (1/8)cos8x]
. . . . . = .(1/128) [5 + 16cos2x - 16cos³2x - 4cos4x - cos8x]
The cosines are all to odd powers.
. . I assume you can do the integration now.
Thanks Soroban!
I found out another way to tackle this question:
Using the rule:
Therefore integral becomes:
[tex]\int_{0}^{\pi/2}cos^6dx-\int_{0}^{\pi/2}cos^8xdx[tex]
Where and
So
When is even
And
Therefore:
Hopefully will look bit more pretty when the LaTex is enabled.
The previous example was when was even.
Using this rule:
etc.
If is even then for example when :
[tex]\frac{7.5.3.1}{8.6.4.2}.\frac{\pi}{2}
Note: factors end with 1 on the top and then add the factor
If is odd then for example when
[tex]\frac{6.4.2}{7.5.3.1}
Note: factors end with 1 on the bottom