hi, I was wondering if someone could show me how to do a problem. I'm studying for a test...and I wanted to be prepared in case something like this came up.

how to you integrate the following...

cosine cubed of x * sin to the fifth of x dx

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- February 12th 2007, 04:20 PMclockinglyIntegrating
hi, I was wondering if someone could show me how to do a problem. I'm studying for a test...and I wanted to be prepared in case something like this came up.

how to you integrate the following...

cosine cubed of x * sin to the fifth of x dx - February 12th 2007, 04:45 PMThePerfectHacker
This happens to be a long integral.

But the first set is to use integration by part,

u=x and v'=sin^5 x

Thus, u'=1

But when you find v you need to find the integral of sin^5 x which you can do by writing it as:

sin^5 x=sin^4 x sin x = (1-cos^2x)^2*sin x

Then use the substitution t=sin x.

But after you do that long integration by parts you will need to do another trigonometric integral like the one above. - February 12th 2007, 10:12 PMSoroban
Hello, clockingly!

Quote:

∫ cos^3(x)·sin^5(x) dx

Then: .[1 - sin^2(x)]·sin^5(x) · cos(x)

And: .[sin^5(x) - sin^7(x)]·cos(x)

We can integrate: .∫ [sin^5(x) - sin^7(x)]·cos(x) dx

. . with the substitution: .u = sin(x)