Posted by Sean on Friday, March 14, 2008 at 9:00pm.
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 don't quite get how you did the first one. Mainly the cos(1/2 x)
Related Questions
calculus help - Is there any good calculus websites out there that teach every ...
Languages - what does .oot derob m'Iþ mean? I have no idea what ...
Calculus - Integrals: When we solve for area under a curve, we must consider ...
CALCULUS - what is the derivative of ln(x-1)? If you are studying the calculus ...
calculus - I'm having trouble with these two problems. If someone can help ...
Calculus (Partial Fractions) - My Calculus class just started Partial Fractions...
CALCULUS 2!!! PLEASE HELP!! - I'm having trouble with this question on arc ...
Hottie - If you want to be banned from this site, keep it up. sorry about my ...
Calculus - Given two problems: Problem A: Tau = Integral(0 to v) v/(q^2 - v^2) ...
calculus - [integrals]2/tsqrt(t^4+25) integrals of two over t square root of t ...
For Further Reading