EXAMPLE 2 Find sin$(7x) cos”7x) dx. SOLUTION We could convert cos?(7x) to 1 - sin?(7x), but...
Evaluate the integral sec 2(7x-2) dx Determine a change of variables from x to u. Choose the correct choice below. O A. u= sec (7x - 2) tan (7x - 2) O B. u=tan (7x-2) O c. u= sec (7x - 2) OD. u = 7x-2 Write the integral in terms of u. sec?(7x+2) dx = S du Evaluate the integral sec2(7x-2) dx = 0
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...
1. Evaluate the indefinite integral sen (2x) – 7 cos(9x) – sec°(3x) dx = 2. Evaluate the indefinite integral | cor(3x) – sec(x) tant(x) + 9 tan(2x) dx = 3. Calculate the indefinite integral using the substitution rule | sec?0 tan*o do =
cos'x dx sin 3x dx 2. an 45 sin cos'xdx 4 sin'xcos'x dr 44 sin'x cos'r dr 6. sin'xcosx dx 8. Jo sin'x cosx dx fa-sin 2x)' dx sin x + cos x dx 10. 9 f sin'z dx cos'x sin'x d 12. 11 sin'x Vcosx dx 14. 13. cot'r sin'x dx 16. cos'x tan'xdx 15 dx sin x dx 18. 17 1-sin x cos x tan'x dx 20. tanx dx 19 sec'x d sec'x dx 22. 21 tan'x secxdx...
Evaluate the integral cos(3x) (1 + 2 sin(3x))\n(1 + 2 sin(3x)) Saint dx
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
Find a solution 6. (e* sin y + tan y)dx + (e* cos y + x(sec y)2)dy = 0.
find a solution 6. (e* sin y + tan y)dx + (e* cos y + x(sec y)2)dy = 0.
Evaluate the integral. 4) S -2x cos 7x dx Integrate the function. dx (x2+36) 3/2 5) S; 5) Express the integrand as a sum of partial fractions and evaluate the integral. 7x - 10 6) S -dx x² . 44 - 12 6)
Please help with each box of the problem. EXAMPLE 5 Evaluate 7x2 - 6x + 16 x3 + 4x dx. Since x3 + 4x = x(x2 + 4) can't be factored further, we SOLUTION write А. Bx + C 7x² - 6x + 16 x(x2 + 4) + Multiplying by x(x2 + 4), we have 7x2 - 6x + 16 = A(x2 + 4) + + Cx + 4A. Equating coefficients, we obtain A + B = 7 C =...