Answer:
Correct integral, third graph
Step-by-step explanation:
Assuming that your answer was 'tan³(θ)/3 + C,' you have the right integral. We would have to solve for the integral using u-substitution. Let's start.
Given : ∫ tan²(θ)sec²(θ)dθ
Applying u-substitution : u = tan(θ),
=> ∫ u²du
Apply the power rule ' ∫ xᵃdx = x^(a+1)/a+1 ' : u^(2+1)/ 2+1
Substitute back u = tan(θ) : tan^2+1(θ)/2+1
Simplify : 1/3tan³(θ)
Hence the integral ' ∫ tan²(θ)sec²(θ)dθ ' = ' 1/3tan³(θ). ' Your solution was rewritten in a different format, but it was the same answer. Now let's move on to the graphing portion. The attachment represents F(θ). f(θ) is an upward facing parabola, so your graph will be the third one.
Answer:

Step-by-step explanation:
Let x, y , and z be the numbers.
Then the geometric sequence is 
Recall that term of a geometric sequence are generally in the form:

This implies that:
a=32 and 
Substitute a=32 and solve for r.


Take the fourth root to get:
![r=\sqrt[4]{\frac{81}{256} }](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B81%7D%7B256%7D%20%7D)

Therefore 


Answer:

Step-by-step explanation:
For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.
and 
Recall the following reciprocal identity:

So, the original expression can be written in terms of only sines and cosines:





Working toward one of the answers provided, this is the tangent function.
The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to
.
A is correct :)
Hope this helps!
Answer:
I'm not positive but I'm pretty sure its infinite
Step-by-step explanation:
idk