A = 21/2 i will write something too just to fill up the space
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:
4: A
Step-by-step explanation:
Mathematically, the average speed over a particular range will be;
f’(b)-f’(a)/(a-b)
In this case; a = 2 and b = 10
So we have the following;
f’(x) = 4t + 1
f’(10) = 4(10) + 1 = 41
f’(2) = 4(2) + 1 = 9
So we have the average speed change as;
(41-9)/(10-2)
= 32/8 = 4
Need to see a picture in order to solve it
