Hey all,

What are the steps to solve the following integral:

kind regards,

Dadon

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- Feb 10th 2007, 01:01 PMdadonIntegration
Hey all,

What are the steps to solve the following integral:

kind regards,

Dadon - Feb 11th 2007, 04:47 AMdadon
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?

- Feb 11th 2007, 03:05 PMSoroban
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.

- Feb 12th 2007, 10:50 AMdadonre:
- Feb 12th 2007, 03:49 PMSoroban
Hello, Dadon!

What part is puzzling you? . . . You just*cube*it . . .

We had: .[½(1 + cos2x)]³ . = . (½)³(1 + cos2x)³

So we have: .(½)³ .= .1/8

Binomial Expansion: .(a + b)³ .= .a³ + 3a²b + 3ab² + b³

So: .(1 + cos2x)³ .= .1 + 3cos2x + 3cos²2x + cos³2x

- Feb 14th 2007, 07:52 AMdadonre:
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. - Feb 14th 2007, 08:12 AMdadon
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