Use the substitution formula to evaluate the integral. 1/2 COS X s dx (5+5 sin x)...
Use a change of variables to evaluate the following indefinite integral. ( (Vx+5) 4 3 dx J2V Determine a change of variables from x to u. Choose the correct answer below. OA. u= (x + 5)^ OB. u= VX +5 OC. Uz OD. u= Write the integral in terms of u. (Vx+5) dx = du 28x Evaluate the integral (Vx+5)* dx=0 2/8 Click to select your answer(s).
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...
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
1. Evaluate the following definite integral using the substitution formula: LI 4 cos(x) sin(x)dr.
QUESTION 18 Use the substitution z = tan(x/2) to evaluate the integral / 3-cos e de ОА. tan-1 ( ✓2 tan 2 --()) +C OB. tan tan +C V2 OC V2 tan 2 Etan () )+c 2 OD. 1 tan tan +C 2 OE. tan V2 tan +C 2
Evaluate the integral using the given substitution. OA. $x*(88 - 4) *dx, u=x2-4 35x25 - 4+0 O B. 15 (85 - 4) + 12 / 3 (x3 - 4) + OD. 13 (x² - 4) + c +C Oc. +C
4. Use an appropriate substitution to evaluate the following integral: 3/4 cos(V1 – x (1 – x dx 0
Evaluate a) integral 0 to pi (dx/5-4 cos x) b) integral 0 to infinity (dx/(1+x^2)^3)
Using reduction formula ***** 13. Evaluate, S sin 5x cos x dx. Also prove that, 53.4 sin mx cos nx dx = 0
EXAMPLE 2 Find sin$(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...