I believe the answer should be 8n - 4.
Answer:
C
Step-by-step explanation:
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
Answer:
a=21.1
Step-by-step explanation:
You can use the given (incorrect) equation and fill in the value of t to find h:
h = 12.5 +9sin(750(3.5)) = 3.68 . . . . feet
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Or, you can use the correct equation, or just your knowledge of revolutions:
h = 12.5 +9sin(750(2π·3.5)) = 12.5 . . . . feet
in 3.5 minutes at 750 revolutions per minute, the propeller makes 2625 full revolutions, so is back where it started — at 12.5 feet above the ground.
