integration??

Question

can some tell me if im doing it right please??

(i) int(4x^3-3\/```x - x^-1/2) dx
(ii) int(4x-3)^3 dx
(iii) int cosx / 3+2sinx dx

(i) int (4x/4 - 9/2 x - 1/2 x^1/2) + k
(ii) int (4(8x^2 - 6^2)^4 +k
(iii) i dont know how to go about doing this one can some help?

thanks bhavinp

Answer 1

With regard to (iii), I'm way out of practice, but I think the usual thing is to integrate by substitution, using the substitutions connected with tan(1/2x).
Worth a try, I hope.

(i) looks right to me.
Wouldn't dismiss (ii).

I'm a graduate mathematician who taught the subject for 10 years, but I'm very rusty at that level now.

Answer 2

(i) I'm having some trouble understanding your problem. Certainly the first term integrates as x^4 + k. I can't read your second term. Now, is the third term (x^(-1))/2 or is it
x^(-1/2)? If the former, the integral is (ln x)/2 + k; if the latter, the integral is (x^(1/2))/(1/2) + k.

(ii) Use the substitution u = 4x - 3, so du = 4dx, or du/4 = dx. The problem becomes int((u^3)du/4) = (1/4)*u^4 + k =
(1/4)*(4x - 3)^4 + k.

(iii) int((cos x)/3 + 2 sin x)dx = (1/3)*int(cos x)dx +
2*int(sin x dx) = (1/3)*sin x + 2*(- cos x) + k.

Answer 3

For the third one and I am very rusty too. Let y =3+2 sinx. Then dy/dx =2cosx. So becomes int(1/2(dy/dx)/y)dx which simplifies to 1/2intdy/y =1/2logey +c =1/2 loge (3 +2 sinx) +c

Answer 4

i dont no what comes after the (4x^3-.... looks like: 3V``` i dont understand this and is der an x missing in the 2nd question?