Thread: Integrating

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

Originally Posted by clockingly

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

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.

Hello, clockingly!

∫ cos^3(x)·sin^5(x) dx

We have: .cos^2(x)·sin^5(x) · cos(x)

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)