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

ID: 2857952 • Letter: I

Question

In this problem you will evaluate

and be given a chance to check your intermediate steps.

-------------------------------------------------------------------

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

-------------------------------------------------------------------

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

x = (8/5)(sec(t)) --> FIRST ANSWER

dx = (8/5)(sect*tant)

And sqrt(25x^2 - 64) --> sqrt(25*64sec^2t/25 - 64) --> sqrt(64sec^2t - 64) ---> sqrt(64)*sqrt(sec^2t - 1)
--> 8*sqrt(tan^2t) --> 8tan(t)

So, integral becomes :

(8/5)(sect*tant) / 8tan(t)

(1/5)sec(t)

So, f(t) = (1/5)(sec(t)) ---> SECOND ANSWER

Integrating :

(1/5)*ln|Sec(t) + tan(t)| ---> THIRD ANSWER

Given x = (8/5)(sec(t))

sec(t) = 5x/8

t = arcsec(5x/8) ---> FOURTH ANSWER

(1/5)*ln|Sec(arcsec(5x/8)) + tan(arcsec(5x/8))|

(1/5)ln|5x/8 + sqrt(25x^2 - 64)/8| + C

So, it becomes :

(1/5)ln| (5x + sqrt(25x^2 - 64)) / 8 | + C ---> FIFTH ANSWER

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.