I worked this out, and the answer is B.) 14x+6
Hope it helped! :3
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
Answer:
w= 9
Step-by-step explanation:

Square both sides:
-4w +61= (w -4)²

Expand:
-4w +61= w² -2(w)(4) +4²
-4w +61= w² -8w +16
Simplify:
w² -8w +16 +4w -61= 0
w² -4w -45= 0
Factorize:
(w -9)(w +5)= 0
w -9= 0 or w +5= 0
w= 9 or w= -5 (reject)
Note:
-5 is rejected since we are only taking the positive value of the square root here. If the negative square root is taken into consideration, then w= -5 would give us -9 on both sides of the equation.
<u>Why </u><u>do </u><u>we </u><u>use </u><u>negative </u><u>square </u><u>root?</u>
When solving an equation such as x²= 4,
we have to consider than squaring any number removes the negative sign i.e., the result of a squared number is always positive.
In the case of x²= 4, x can be 2 or -2. Thus, whenever we introduce a square root, a '±' must be used.
However, back to our question, we did not introduce the square root so we should not consider the negative square root value.