ANSWER:
Let t = logtan[x/2]
⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx
⇒dt = 1/2 cos² x/2 × cot x/2dx
⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx
⇒dt = 1/2 cosx/2 / sin x/2 dx
⇒dt = 1/sinxdx
⇒dt = cosecxdx
Putting it in the integration we get,
∫cosecx / log tan(x/2)dx
= ∫dt/t
= log∣t∣+c
= log∣logtan x/2∣+c where t = logtan x/2
Answer:
f'(1) = 2
Step-by-step explanation:
f(x) = 2x^2 -2x +3
Take the derivative
f'(x) = 2 * 2x - 2 * 1
f'(x) = 4x -2
f'(1) = 4(1) -2
=4-2
=2
Answer:
do the test then
Step-by-step explanation:
do the test then
You can find the y-intercept by plugging in zero for x and solving. We know that the value will always be zero when the point is on the y-axis, which is the y-intercept.
3(0) -4y = 16
-4y = 16
divide both sides by -4
y = -4
Letter A
