O to the power of 2 i think
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
Step-by-step explanation:
p^2-3p(3/2)^2=8+(3/2)^2
p^2-(3/2)^2=8+9/4
√(p-3÷2)^2=√(8+9÷4) take LCM and cancel out the square and square root
p-3÷2=√(32+9÷4)
p=(3+_√41÷2)
Answer:
hold on i'm abt to answer it
Step-by-step explanation:
C is the answer it’s supposed to be X+33
