Hey all,
What are the steps to solve the following integral:
$\displaystyle \int_{0}^{\pi/2}sin^2xcos^6xdx$
kind regards,
Dadon
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:
$\displaystyle \int_{0}^{\pi/2}sin^2xcos^6xdx$
Using the rule:$\displaystyle sin^2x=1-cos^2x$
Therefore integral becomes:
$\displaystyle \int_{0}^{\pi/2}(1-cos^2x)cos^6xdx$
[tex]\int_{0}^{\pi/2}cos^6dx-\int_{0}^{\pi/2}cos^8xdx[tex]
Where $\displaystyle I_6=\int_{0}^{\pi/2}cos^6dx$ and $\displaystyle I_8=\int_{0}^{\pi/2}cos^8xdx$
So $\displaystyle I_6-I_8$
When $\displaystyle I_n$ is even
$\displaystyle I_6= \frac{(6-1)}{6}.\frac{(6-3)}{(6-2)}.\frac{(6-5)}{(6-4)}.\frac{\pi}{2}$
$\displaystyle I_6=\frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi} {2} = \frac{15\pi}{96}$
And
$\displaystyle I_8=\frac{(8-1)}{8}.\frac{(8-3)}{(8-2)}.\frac{(8-5)}{(8-4)}.\frac{(8-7)}{(8-6)}.\frac{\pi}{2}$
$\displaystyle I_8=\frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2 }.\frac{\pi}{2} = \frac{105\pi}{768}$
Therefore:
$\displaystyle I_6-I_8 = \frac{15\pi}{96}-\frac{105\pi}{768}$
$\displaystyle \frac{5\pi}{256}$
Hopefully will look bit more pretty when the LaTex is enabled.
The previous example was when $\displaystyle I_n$ was even.
Using this rule:
$\displaystyle \frac{(n-1)(n-3)(n-5)...}{n(n-2)(n-4)(n-6)...}$ etc.
If $\displaystyle n$ is even then for example when $\displaystyle n = 8$:
[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 $\displaystyle \frac{\pi}{2}$
If $\displaystyle n$ is odd then for example when $\displaystyle n=7$
[tex]\frac{6.4.2}{7.5.3.1}
Note: factors end with 1 on the bottom