Math

posted by .

Evaluate the Integral

pi
/ = (3cos^2x-1) (sinx) dx=0
0


I am really confused with this, please help!
** used / for integral, not sure if there is a way to type it!

  • Math -

    ∫[0,pi] (3cos^2x-1) (sinx) dx
    let u = cosx, then du = -sinx dx
    now you have
    ∫[1,-1] -(3u^2-1) du
    = u^3 - u [-1,1]
    = 0

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Integral

    That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten integral …
  2. calc asap!

    can you help me get started on this integral by parts?
  3. Calculus

    integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is improper or involves improper integral *infinite limit of integration (b) determine whether the integral converges or diverges converges?
  4. Calculus

    Use the symmetry of the graphs of the sine and cosine functions as an aid in evaluating each definite integral. (a) Integral of sinx*dx from -pi/4 to pi/4 (b) Integral of cosx*dx from -pi/4 to pi/4 (c) Integral of cosx*dx from -pi/2 …
  5. Math(Please check)

    evaluate the integral integral of 3 to 2 x/(x^2-2)^2 dx u=x^2-2 du=2x dx 1/2 du = x dx integral of 1/u^2 du -1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other -1/(3^2-2) - (-1/(2^2-2) Is this correct?
  6. Math

    1) evaluate: integral of (3x^2 + 5x - 1 + 3/x) dx x^3/3 + x^2/5 - x Is this correct?
  7. Calculus

    Evaluate the following integral integral 1 = a and b = 4 of sinx dx/(1+cos)^2 u = cosx du = -sinx dx so from here I don't know if I can do: -1 du = sinx dx or 1/sin du = x dx
  8. Math

    Let A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4. 1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A. My Work: for x: integral[4,1] …
  9. Calculus Question! ASAP!

    Hello! I have this problem: x(dx)/sqrt(9-x^2) I was wondering why I can't use trig substitution and substitute sqrt(9-x^2) for sqrt(1-sec^2) and having: integral x = 3sin(theta) dx = 3cos(theta)d(theata) integral 3sin(theta)(3cos(theta))/3cos(theta) …
  10. Calculus

    Evaluate the integral. 1/2 integral e^(t/2) (I'm not sure what the 1/2 on the left of the integral symbol means.)

More Similar Questions