Evaluate the following integral
integral 1 = a and b = 4 of
sinx dx/(1+cos)^2
u = cosx
du = -sinx dx
so from here I don't know if I can do:
-1 du = sinx dx
or 1/sin du = x dx
u = 1 + cos x
du = -sin x dx
so we have
-du/u
so I tried again and I got this, don't know if it's good:(I also put the wrong a and b)
integral 0 = a and b = pi/3 of
sinx dx/(1+cos)^2
u = cosx
du= -sinxdx
-1 du = sinx dx
so: integral -1 du/(1+u)^2
= 1/(u+1)
if x = 0, then u = 1
if x = pi/3, then u = 1
so 1/1+1 - 1/1+1 = 0
Is this correct?
-du/(u)^2
-du/u^2 ---> -1/u + c
ok Damon, so if you integrate what you have me, you get 1/u no? Because 1/u^2 = -1/u, so -1/u^2 = 1/u?
so
-1/(1+cos x) + c
Yes, you have it.
awesome, thanks man!
You are welcome.
To evaluate the integral ∫[a,b] sin(x)dx/(1+cos(x))^2, you made a good choice by letting u = cos(x), which leads to du = -sin(x)dx. However, it seems like you're getting a bit confused with the manipulation of the differential terms.
Let's start from the beginning:
1. Start with the substitution u = cos(x), which gives you du = -sin(x)dx.
2. Rewrite the original integral in terms of u:
∫[a,b] sin(x)dx/(1+cos(x))^2 = ∫[a,b] -du/(1+u)^2
Please note that the expression sin(x)dx has been replaced by -du, and (1+cos(x))^2 has been replaced by (1+u)^2.
3. Now you can evaluate the integral:
∫[a,b] -du/(1+u)^2 = - ∫[a,b] du/(1+u)^2
To evaluate this integral, we can use the power rule for integration. According to the power rule, the integral of du/(1+u)^2 is -1/(1+u) + C, where C is the constant of integration.
Thus, the definite integral is:
- ∫[a,b] du/(1+u)^2 = -[-1/(1+u)] evaluated from a to b
= -[-1/(1+u)] evaluated from a to b
= -[-1/(1+cos(x))] evaluated from a to b
So the final result is:
-[-1/(1+cos(b))] - [-1/(1+cos(a))]
Therefore, the value of the original integral is:
[-1/(1+cos(a))] - [-1/(1+cos(b))]