1. ## Integration

Hey all,

What are the steps to solve the following integral:

$\int_{0}^{\pi/2}sin^2xcos^6xdx$

kind regards,

2. 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?

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.

4. ## re:

Originally Posted by Soroban

. . . = .(1/8) (1 + 3cos2x + 3cos²2x + cos³2x)
Hello Soroban,

Could you please exlain how you got the line in the quote.

Thank you

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

6. ## re:

Thanks Soroban!

I found out another way to tackle this question:

$\int_{0}^{\pi/2}sin^2xcos^6xdx$

Using the rule: $sin^2x=1-cos^2x$

Therefore integral becomes:
$\int_{0}^{\pi/2}(1-cos^2x)cos^6xdx$

[tex]\int_{0}^{\pi/2}cos^6dx-\int_{0}^{\pi/2}cos^8xdx[tex]

Where $I_6=\int_{0}^{\pi/2}cos^6dx$ and $I_8=\int_{0}^{\pi/2}cos^8xdx$

So $I_6-I_8$

When $I_n$ is even

$I_6= \frac{(6-1)}{6}.\frac{(6-3)}{(6-2)}.\frac{(6-5)}{(6-4)}.\frac{\pi}{2}$
$I_6=\frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi} {2} = \frac{15\pi}{96}$

And
$I_8=\frac{(8-1)}{8}.\frac{(8-3)}{(8-2)}.\frac{(8-5)}{(8-4)}.\frac{(8-7)}{(8-6)}.\frac{\pi}{2}$
$I_8=\frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2 }.\frac{\pi}{2} = \frac{105\pi}{768}$

Therefore:
$I_6-I_8 = \frac{15\pi}{96}-\frac{105\pi}{768}$

$\frac{5\pi}{256}$

Hopefully will look bit more pretty when the LaTex is enabled.

7. The previous example was when $I_n$ was even.

Using this rule:
$\frac{(n-1)(n-3)(n-5)...}{n(n-2)(n-4)(n-6)...}$ etc.

If $n$ is even then for example when $n = 8$:

[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 $\frac{\pi}{2}$

If $n$ is odd then for example when $n=7$

[tex]\frac{6.4.2}{7.5.3.1}

Note: factors end with 1 on the bottom

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