In this problem you will evaluate and be given a chance to check your intermedia

ID: 2857941 • Letter: I

Question

In this problem you will evaluate

and be given a chance to check your intermediate steps.

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Answer:
Do not include an arbitrary constant "+ C" in your answer.

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Answer: +C
Do not include an arbitrary constant "+ C" in your answer.

x = 5sin(t) --> FIRST ANSWER

dx = 5cos(t)*dt

sqtr(25 - x^2) --> sqrt(25 - 25sin^2t) --> sqrt(25)*sqrt(1 - sin^2t) --> 5*sqrt(cos^2t) --> 5cost

So, problem becomes :

5cost * 5cost*dt

25cos^2(t)*dt

f(t) = 25 * cos^2(t) --> SECOND ANSWER

(integral)25*cos^2(t)*dt

25 * (integral) [1 + cos(2t)]/2 * dt

25*[t/2 + sin(2t)/4] + C

25t/2 + 25sin(2t)/4 + C ---> THIRD ANSWER

x = 5sin(t)

x/5 = sin(t)

t = arcsin(x/5) --> FOURTH ANSWER

Plug this back into the intyegral solution.....

25t/2 + 25sin(2t)/4 + C
becomes

(25/2)arcsin(x/5) + (25/4)sin(2arcsin(x/5) ) + C

(25/2)arcsin(x/5) + (25/2)sin(arcsin(x/5))cos(arcsin(x/5)) + C

(25/2)arcsin(x/5) + (25/2)(x/5)(sqrt(25 - x^2)/5) + C

(25/2)arcsin(x/5) + (x/2)*sqrt(25 - x^2) + C ---> FIFTH ANSWER

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