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!
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)
No worries! I'm here to help you with your math questions. Let's tackle each problem one at a time.
1) To find the integral of [2* sqrt((1+cosx)/2)], you can start by pulling the constant multiplier (2) out of the integral, as you mentioned:
∫ [2* sqrt((1+cosx)/2)] dx
= 2 * ∫ sqrt((1+cosx)/2) dx
Now, let's focus on the integral ∫ sqrt((1+cosx)/2) dx. To simplify this, we can use a trigonometric identity:
cos^2(x/2) = (1 + cosx)/2
Rearranging the identity, we have:
(1 + cosx) = 2 * cos^2(x/2)
Substituting this into our integral, we get:
2 * ∫ sqrt(2 * cos^2(x/2)/2) dx
Simplifying,
2 * ∫ cos(x/2) dx
Now, you can solve this integral using the substitution method. Let u = x/2, then du = (1/2) dx. Rearranging, we have dx = 2 du.
Now substitute u and du into the integral:
2 * ∫ cos(u) * 2 du
= 4 ∫ cos(u) du
= 4 sin(u) + C
Finally, substitute back u = x/2:
= 4 sin(x/2) + C
So, the integral of [2* sqrt((1+cosx)/2)] is 4 sin(x/2) + C, where C is the constant of integration.
Now let's move on to the second problem.
2) To find the integral of [(sec(x))^4], there is a neat trick you can use involving the identity:
sec^2(x) = 1 + tan^2(x)
First, rewrite [(sec(x))^4] as [(sec^2(x))^2]:
∫ [(sec(x))^4] dx
= ∫ [(sec^2(x))^2] dx
Now, use the identity sec^2(x) = 1 + tan^2(x):
= ∫ [(1 + tan^2(x))^2] dx
Expanding the square inside the integral, we get:
= ∫ [1 + 2tan^2(x) + tan^4(x)] dx
Now, this integral can be solved by separating it into three separate integrals:
∫ [1 + 2tan^2(x) + tan^4(x)] dx = ∫ dx + 2 ∫ tan^2(x) dx + ∫ tan^4(x) dx
The integral of dx is simply x + C (constant of integration).
To solve the integrals of tan^2(x) and tan^4(x), you can use trigonometric substitution or integration by parts. Both methods are slightly more involved, but they will give you the final results:
∫ tan^2(x) dx = tan(x) - x + C1
∫ tan^4(x) dx = (tan^3(x))/3 - x + C2
Now, you can combine the results to find the final integral of [(sec(x))^4]:
∫ [(sec(x))^4] dx = x + 2(tan(x) - x) + (tan^3(x))/3 - x + C
= -2x + 2tan(x) + (tan^3(x))/3 + C3
So, the integral of [(sec(x))^4] is -2x + 2tan(x) + (tan^3(x))/3 + C3, where C3 is the constant of integration.
As for your request for math tutorials, there are plenty of online resources available. Khan Academy (www.khanacademy.org) is a popular platform that offers math tutorials and exercises for various topics. Additionally, you can find math tutorials on websites like YouTube or math-specific forums. Just search for the specific topic you're interested in, and you're likely to find a wealth of resources to help you understand and practice.