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
The answer is c just plug in the value for C and D
Yes, they are all per something so yes
Answer:
Step-by-step explanation:
