Answer:
½ sec²(x) + ln(|cos(x)|) + C
Step-by-step explanation:
∫ tan³(x) dx
∫ tan²(x) tan(x) dx
∫ (sec²(x) − 1) tan(x) dx
∫ (sec²(x) tan(x) − tan(x)) dx
∫ sec²(x) tan(x) dx − ∫ tan(x) dx
For the first integral, if u = sec(x), then du = sec(x) tan(x) dx.
∫ u du = ½ u² + C
Substituting back:
½ sec²(x) + C
For the second integral, tan(x) = sin(x) / cos(x). If u = cos(x), then du = -sin(x) dx.
∫ -du / u = -ln(u) + C
Substituting back:
-ln(|cos(x)|) + C
Therefore, the total integral is:
½ sec²(x) + ln(|cos(x)|) + C
Function A because in one hour it gets up by 4 but in function B it gets up by only 2
Answer: t = 33
Step-by-step explanation:
t + 5 = 38
-5 -5 subtract 5 from both sides in order to isolate t
t = 33
(y-y1) = m(x-x1)
y-2 = 5(x-6)
Therefore, point (x1,y1) lies on the line of equation
Therefore, point (6,2) lies on the line with the point slope equation y-2=5(x-6)
all you have to do is add 2.09 to itself until it gets to 22.5
2.09 x 10 = 20.9
2.09 x 11 = 22.99
so, 2.09 goes into 22.5 ten times
