st sin t dt = -(A) - tan s (B) tan1s tan-ls (C) tan-1s (D) T- (E) None
Question The value of the integrale sint t )dt Select one: O a. equals to O O b. equals to 1. O ö C. equals to 0. d. equals to o e. does not exist.
Solve the initial value problem. ds dt = cost – sint, s (7) = 3 NOTE: This question is bonus, worth 5 points. Os=sint + cost + 2 8 = sint + cost +4 None of them 8 = sint - cost + 2 8=2 sint + 1
Use the given information to find (a) sin (s +t), (b) tan (s + t), and (c) the quadrant of s+t. 12 13 and sint = 5 s and t in quadrant III 4 COS S =- (a) sin (s+t)=0 (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) (b) tan (s+t) = 0 (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) (c) What is...
If •dt, then Vtant g'(x)= a) cos x 1 w b) tan x c) 2v tan x d) cosx
Find the general indefinite integral.
15. 14 1 - sint -dt sint Answer sin 2.c 16. dc sin 3
Consider: S x2-yds, C: r(t) = (e"? 2, 1+e'), te[0,2] Which one of the following "regular" integrals represents the above line integral. dt O a. Ob. V 4 dt 0 S'Vertel dat O d.o Question 8 10 point Consider: | <x?,v/dr, C: r(t) = (sint, cost), te[0,1] Which one of the following "regular" integrals represents the above line integral. S". cost sint - cost sint dt O a. o П 1 sin2tdt 0 s "cost sin’t + cost sint dt...
1. (a) We want to develop a method for calculating the function sint dt f)-inf t 0 for small or moderately small values of x. This is a special function called the "sine integral", and it is related to another special function called the "exponential integral". It arises in diffraction problems. Derive a Taylor-series expression for f(x), and give an upper bound for the error when the series is terminated after the n-th order term. [HINT: (-1)"*z ? + R...
Find f. f'(t) = 2t - 4 sint, (0)= 5 Select one: a. f(t) = 2t - 3 sint +5 O b. f(t)= +2 +4 cost +1 c. f(t)=12 - 3 cost-5 d. f(t) = x2 +3 cost e. None of these
(1 point) Evaluate the following: b.(3+e 2)8(t - 9) dt- (3 + e 2t)<s(t) dt-