Calculus homework?

Question

1. Find the length of the given graph over the indicated interval. r= 5 ( 1 + cos (delta)). Interval: 0< delta < 2 pie.

2. Convert the given rectangular equation to polar form:
4 ((x)^2) - 5 ((y)^2) - 36 y - 36 = 0.

3. lim (-x ln x)=?
*note as limit approaches 0 from the positive side.

4. the area bounded by r^2 =sin (delta).

5. The area bounded by the small loop of r = 2-4sin(delta) .

6. the integral arctan (2x) dx, between the interval (0, 1/2).

7. if f(x) = 2 ^(sin(^(2)) (x)), then f' (pie/4) equals ?

Answer 1

1) =5Int(0,2pi)sqrt[1+cost)^2+sin^2 t) ]dt=10 Int sqrt(1+cos t)dt =
=20 sqrt(1+cos t)*tan(t/2) Taken between (0,2pi)
2) =4r^2 cos^2 t-5r^2sin^2t -36r sin t -36=0
3) lim-x ln x = - lim ln x/1/x and by L´Hôpital = lim1/x/(1/x^2)=limx=0
4)A = 1/2 Int sin t dt (0,pi) = -1/2cost) = 1

6)call 2x = z dx=dz/2
1/2 Int arctanz dz m ( 0,1) = 1/2(pi/4-1/2ln2)

7)f´(x) = 2^sin^2x *ln2 * 2sinx cos x= 2^1/2*ln2 at x=pi/4
5) r= 4( 1/2+cost) and the area is 2(2pi-3sqrt3)

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