Help solve the following integral: 1/2∫ (4 + 4cosθ)^2 - (6)^2 dθ?

Question

and the upper and lower limits are θ = pi/3, -pi/3. I try to solve as follows..:

1/2∫ (4 + 4cosθ)^2 - (6)^2 dθ

1/2∫ (16 + 32cosθ + 16cos^2θ - 36 dθ <--Expand

1/2∫ -20 + 32cosθ + 16 1/2 (1 + cos2θ) dθ <--Trig identity

1/2∫ -20 + 32cosθ + 8 (1 + cos2θ) dθ <--cancel 16 and 1/2

1/2 [ -20θ + 32sinθ + 8θ + 4sin2θ]

1/2[-12θ + 32sinθ + 4sin2θ], then I'd replace the θ with the upper and lower limits, however, I get the wrong answer. Where is my mistake? Thanks!

Answer 1

4+5= 9

Answer 2

No mistake in what you've written above. The given integral does evaluate to 18√3 - 4π ≈ 18.61, which is what your expression gives for the specified limits. (If you got something else, it's your substitution or simplification which is in error.)

If this is definitely not the right answer, then your integral is not correct. Without knowing the problem, though, I can't tell you what the integral should be.

Be aware that it's not unknown for some of the provided answers to be wrong.

Answer 3

whats up there! here is the respond. ? 3/(x+a million)^2+4 dx --> Write the subject. 3? a million/(x+a million)^2+4 dx --> Use the valuables ? cf(x) dx=c? f(x). u=x+a million, then du=a million --> Use substitution, permit x+a million equivalent u, then the by-made from u is the comparable because of the fact the by-made from x+a million, it somewhat is a million. 3? a million/u^2+4 du --> substitute u for x+a million. 3? a million/u^2+2^2 du --> Rewrite 4 as 2^2. 3(a million/2arctan(u/2))+C --> Use the valuables ? a million/(u^2+a^2) du=a million/a*arctan(u/a)+C. 3(a million/2arctan((x+a million)/2))+C --> substitute x+a million for u. 3/2arctan((x+a million)/2)+C Distribute 3 into the above expression. So the respond is 3/2arctan((x+a million)/2)+C. desire it helps!

Answer 4

1/2[ -12(2pi/3)+32(sqrt(3) )+4(sqrt(3) ) ]
=1/2[ -8pi+36sqrt(3) ]
= -4pi +18 sqrt(3)