math

posted by .

how would i find the integral of sin (2t+1)/cos^2(2t+1) dt

  • math -

    let u = (2t+1)
    du = 2 dt so dt = .5 du
    so
    .5 integral du sin u/cos^2u

    try 1/cos u
    d/du (1/cos u) = -sin u/cos^2 u

    You can take it from there :)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. 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 …
  2. algebra

    Can someone please help me do this problem?
  3. 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 again. …
  4. 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 …
  5. calc

    find the area between the x-axis and the graph of the given function over the given interval: y = sqrt(9-x^2) over [-3,3] you need to do integration from -3 to 3. First you find the anti-derivative when you find the anti-derivative …
  6. TRIG!

    Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …
  7. 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 …
  8. 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: …
  9. Integral Help

    I need to find the integral of (sin x)/ cos^3 x I let u= cos x, then got -du= sin x (Is this right correct?
  10. 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 cost(t)e^(it)dt …

More Similar Questions