Answer:
(e) csc x − cot x − ln(1 + cos x) + C
(c) 0
Step-by-step explanation:
(e) ∫ (1 + sin x) / (1 + cos x) dx
Split the integral.
∫ 1 / (1 + cos x) dx + ∫ sin x / (1 + cos x) dx
Multiply top and bottom of first integral by the conjugate, 1 − cos x.
∫ (1 − cos x) / (1 − cos²x) dx + ∫ sin x / (1 + cos x) dx
Pythagorean identity.
∫ (1 − cos x) / (sin²x) dx + ∫ sin x / (1 + cos x) dx
Divide.
∫ (csc²x − cot x csc x) dx + ∫ sin x / (1 + cos x) dx
Integrate.
csc x − cot x − ln(1 + cos x) + C
(c) ∫₋₇⁷ erf(x) dx
= ∫₋₇⁰ erf(x) dx + ∫₀⁷ erf(x) dx
The error function is odd (erf(-x) = -erf(x)), so:
= -∫₀⁷ erf(x) dx + ∫₀⁷ erf(x) dx
= 0
A storage tank. I hope this helps ^^
Answer:
400mL/3l
Is the correct answer
Step-by-step explanation:
Integrate the force field along the given path (call it <em>C</em>):


