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.
I noticed that this equation is in slope-intercept form (y = mx + b). M = the slope of the line, B = the y-intercept.
If the line is translated (fancy, fancy, talk for MOVED) up. The Y-intercept would change positive (because it is moved up) by four.
4 + 4 equals...well, 8.
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Complementary angles add up to 90 so
angle + 47 = 90
angle = 90 - 47
angle = 43 so A
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