1) use the formula:
cos(2x) = 2 cos^2(x) - 1
You can use this formula to derive a formula for cos(1/2 x) in terms of
cos(x). You can then simplify the square root using that formula.
2) sec^4 (x) = 1/cos^4(x)
You can use a reduction formula, which is easier to derive for positive powers of cos,, you just apply that backwards for this case.
Notation cos = c, sin = s:
c^n = c^(n-2)c^2 = c^(n-2)[1-s^2] =
c^(n-2) - c^(n-2)s^2
So the integral of c^n is the integral of c^(n-2) minus the integral of
c^(n-2)s^2. This latter integral can be written up to a minus sign:
Int of c^(n-2)sdc = ( let's do a partial integration) =
Integral of s d[c^(n-1)/(n-1)] =
Integral of d[sc^(n-1)/(n-1)] -
Integral of c^(n-1)/(n-1)ds =
sc^(n-1)/(n-1) -
Integral of c^(n)/(n-1) dx
So, we have expressed the integral of cos^n in terms of the integral of
cos^(n-2), a trignometric function and cos^n again. You can bring that later integral of cos^n back to the other side of this expression and solve for the integral of cos^n in terms of the integral of cos^(n-2).
You can then verify that this equation is also valid for negative n (that's is trivial). So, you can use it to express the integral of 1/c^4 in terms of the integral of 1/c^2, that latter integral is of course, the tan function.
I know I've probably flooded the site or something (Sorry if I have), it's just that these problems aren't similar to anything I've reviewed in class. I've got two more questions, then maybe 1 or 2 more coming. That's all I can see in the future, anyhow.
1) Integral of [2* sqrt((1+cosx)/2)].
All I can see is that I take the 2 out of the integral, and put it out front.
2) Integral of [(sec(x))^4].
Here I'm completely stumped. I'm not sure how to start, either.
I'd really appreciate help on these two problems. They've really got me stuck.
P.S. Also, if anyone could supply a link of math tutorials of some sort, please do!
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I don't quite get how you did the first one. Mainly the cos(1/2 x)
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