what is the integral of sin^5(6x)dx

  1. 👍 0
  2. 👎 0
  3. 👁 99
asked by walter
  1. first, note that if u = 6x, you have

    1/6 ∫ sin^5 u du
    = 1/6 ∫sin^4(u) sinu du
    = -1/6 ∫(1-cos^2 u)^2 (-sinu du)
    If v = cos u, then dv = -sinu du, and you have

    -1/6 ∫(1-v^2)^2 dv

    just expand that and it's a regular old polynomial.

    1. 👍 0
    2. 👎 0
    posted by Steve

Respond to this Question

First Name

Your Response

Similar Questions

  1. calc

    find integral using table of integrals ) integral sin^4xdx this the formula i used integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n integral sin^n-2 using the formula this is what i got: integral sin^4xdx=-1/4sin^3xcosx+3/4 integral

    asked by tom on February 20, 2011
  2. Integration by Parts

    integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral

    asked by Ashley on April 16, 2015
  3. 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

    asked by drwls on February 20, 2007
  4. trig integration

    s- integral endpoints are 0 and pi/2 i need to find the integral of sin^2 (2x) dx. i know that the answer is pi/4, but im not sure how to get to it. i know: s sin^2(2x)dx= 1/2 [1-cos (4x)] dx, but then i'm confused. The indefinite

    asked by christine on February 18, 2007
  5. Maths

    Question : Integrate [x/(1+(sin a*sin x))] from 0 to pi My first thought was to apply integrate f(x) dx= f(a-x) dx method Which simplified the integral into; 2I = integrate [pi/(1+(sin a*sin x))] dx , cancelling out x Then I made

    asked by Ashley on March 18, 2019
  1. math

    I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up the integral and proceed to do integration by parts twice, but it keeps working out to 0=0 or sin(t)=sin(t). How am I supposed to approach it? integral

    asked by chuck on December 3, 2009
  2. trig integration

    i'm having trouble evaluating the integral at pi/2 and 0. i know: s (at pi/2 and 0) sin^2 (2x)dx= s 1/2[1-cos(2x)]dx= s 1/2(x-sin(4x))dx= (x/2)- 1/8[sin (4x)] i don't understand how you get pi/4 You made a few mistakes, check

    asked by christine on February 18, 2007
  3. Calculus

    Evaluate the integral. S= integral sign I= absolute value S ((cos x)/(2 + sin x))dx Not sure if I'm doing this right: u= 2 + sin x du= 0 + cos x dx = S du/u = ln IuI + C = ln I 2 + sin x I + C = ln (2 + sin x) + C Another problem:

    asked by CMM on March 8, 2011
  4. Math/Calculus

    How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Just like you did x^2 exp(x) below. Also partial integration is not the easiest way to do this integral. You can also use this

    asked by COFFEE on May 28, 2007
  5. Calculus

    I am trying to find the integral of e^(6x)sin(7x). Apparently, the answer is (6e^(6x)sin(7x) - 7e^(6x)cos(7x))/85) + C and achieving the answer is mostly understandable since it involves integrating the function in parts... that

    asked by Lindsey on October 7, 2018
  6. Calculus

    I have two questions, because I'm preparing for a math test on monday. 1. Use the fundamental theorem of calculus to find the derivative: (d/dt) the integral over [0, cos t] of (3/5-(u^2))du I have a feeling I will be able to find

    asked by Jennifer on May 9, 2009

More Similar Questions