Calculus

posted by .

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:

S ((sin (ln x))/(x)) dx

I don't even know what to put as u and du?

Please help with explanations.

  • Calculus -

    The first problem is correct.
    As for the second problem, take:
    u = ln x
    du = (1/x) dx

    So:
    S (ln u) du = ?
    Try it yourself

  • Calculus -

    Why is it S (ln u) du = ?
    What about the sin? And to multiply them are you reciprocating?

  • Calculus -

    Oh sorry. It should be:
    S sin(u) du

  • Calculus -

    So I get this:

    = ln Isin(u)I + C
    = ln Isin(ln x)I + C
    = ln (sin(ln x)) + C
    OR
    = ln (sin(1/x)) + C

    Is it correct or incorrect?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calc

    d/dx integral from o to x of function cos(2*pi*x) du is first i do the integral and i find the derivative right. by the fundamental theorem of calculus, if there is an integral from o to x, don't i just plug the x in the function. …
  2. 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 …
  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. 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 …
  6. Calculus

    1.evaluate (integral sign)x cos 3x dx A.1/6 x^2 sin 3x + C B.1/3 x sin 3x -1/2 sin 3x +C C.1/3 x sin 3x +1/9 cos 3x +C << my choice. D. 1/2 x^2 +1/18 sin^2 3x +C 2.evaluate (integral sign)xe(power of 2x)dx A.1/6 x^2 e(to the …
  7. 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?
  8. 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 …
  9. Calculus

    Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused …
  10. Calculus

    Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused …

More Similar Questions