QUESTION 3 Evaluate the integral by using multiple substitutions. SV1 1 + sin2 (x-7) sin (x-7)...
need only the answer Evaluate the integral by using multiple substitutions. dx 313x2 – 2) sin(x3 2x) cos(x3 - 2x) O 2 sin(x3 - 2x) + C 15 sin4 (x3 - 2x) + c o cos6 (3x2)+C o į sin® (x3 - 2x)+ c
1. a) Substitute u = sin(x) to evaluate sin^2(x) cos^3(x) dx. [trig identity sin2(x)+cos2(x) = 1]. b) Find the antiderivatives: i) sin(2x) dx ii) (cos(4x)+3x^2) dx
Evaluate the following integral. 1/2 7 sin ?x -dx 1 + cos x 0 1/2 7 sin 2x dx = V1 + cos x 0 Score: 0 of 1 pt 1 of 10 (0 complete) HW Score: 0%, 0 of 10 pts 8.7.1 A Question Help The integral in this exercise converges. Evaluate the integral without using a table. dx x +49 0 dx X2 +49 (Type an exact answer, using a as needed.) 0
4. Determine the integral which computes the arc length of the curve y = sin(x) with 0 < x <. TT A '1 + sin2(a)dx so $." .TT B 1 + cos2(x)dx С [* V1 – cos? (7)dx D| None of the above.
Tutorial Exercise Evaluate the integral using the substitution rule. sin(x) 1/3 1* dx cos(x) Step 1 of 4 To integrate using substitution, choose u to be some function in the integrand whose derivative (or some constant multiple of whose derivative) is a factor of the integrand. Rewriting a quotient as a product can help to identify u and its derivative. 70/3 1." sin(x) dx = L" (cos(x) since) dx cos?(X) Notice that do (cos(x)) = and this derivative is a...
1P Question 3 1 Evaluate the double integral: SS sin?(x) dx2 7 o (+ cos(2x)) 0} (x2 + cosº (x)) No answer text provided. 0}(– cos(2x)) 0} (x + 2 cos(2x)) NE Previous
1. Evaluate the following indefinite integral using the given trigonometric substitutions, = sin e V1 - 22 = cose U
Evaluate the integral using the given substitution. ſ Vacos (x32-7) dx, u = x 3/2_, o x312 sin 2(x3/2 - 7)+ C o sinᵒ (8312.7) + C a o 1/3 (4x) sin (x3/2 -7)+C o $(x3/2 - 7) + sin 2(x3/2 - 7) + C
(3) Evaluate the indefinite integral. ſtan(x) + cos2 (2) dx cos(2)
EXAMPLE 2 Find sin$(7x) cos”7x) dx. SOLUTION We could convert cos?(7x) to 1 - sin?(7x), but we would be left with an expression in terms of sin(7x) with no extra cos(7x) factor. Instead, we separate a single sine factor and rewrite the remaining sin" (7x) factor in terms of cos(7x): sin'(7x) cos”(7x) = (sinº(7x))2 cos(7x) sin(7x) = (1 - Cos?(7x))2 cos?(7x) sin(7x). in (7x) cos?(7x) and ich is which? Substituting u = cos(7x), we have du = -sin (3x) X...