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Math Help - Integration

  1. #1
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    Integration

    Hi I got a few probs with some integration questions such as

    Integration by parts:

    Limits [0,2] xsinh x dx
    =======
    Change of variable rule

    Limits [-r,r] (r^2-x^2)^1/2 dx where x =rsint r>0
    =======
    and

    Limits ["pie"/4, "pie"/2] (dx/sin x) using t = tan x/2)

    Any kind of help is well appreciated

    Thx
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeloAngelo View Post
    Hi I got a few probs with some integration questions such as

    Integration by parts:

    Limits [0,2] xsinh x dx
    \int _0^2 x~sinh(x)~dx

    Use integration by parts. u = x and v = sinh(x).


    Quote Originally Posted by NeloAngelo View Post
    Change of variable rule

    Limits [-r,r] (r^2-x^2)^1/2 dx where x =rsint r>0
    x = r~sin(t) \implies dx = r~cos(t)~dt

    So
    \int (r^2 - x^2)^{1/2}~dx = \int (r^2 - r^2~sin^2(t))^{1/2}~(r~cos(t)~dt)

    = \int r^2 (1 - sin^2(t))^{1/2}~cos(t)~dt

    = \int r^2 cos^2(t)~dt

    Now for the integration limits.
    x = -r \implies -r = r~sin(t) \implies sin(t) = -1 \implies t = -\frac{pi}{2}

    x = r \implies -r = r~sin(t) \implies sin(t) = 1 \implies t = \frac{pi}{2}

    So the final form for the integral is
    r^2 \int_{-\pi/2}^{\pi/2} cos^2(t)~dt

    -Dan
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by NeloAngelo View Post
    Change of variable rule

    Limits [-r,r] (r^2-x^2)^1/2 dx where x =rsint r>0
    It's a semicircle whose radius is r, hence its value is \frac\pi2r^2.

    Quote Originally Posted by topsquark View Post
    So the final form for the integral is
    r^2 \int_{-\pi/2}^{\pi/2} cos^2(t)~dt
    Having this, don't even think to replace the double angle formula for cosine.

    First, note that this function is even, so the integral can be rewritten as 2r^2\int_0^{\pi /2} {\cos ^2 t\,dt} . Besides \int_0^{\pi /2} {f(\sin x)\,dx}  = \int_0^{\pi /2} {f(\cos x)\,dx} whose proof is easy, so \int_0^{\pi /2} {\cos ^2 t\,dt}  = \int_0^{\pi /2} {\sin ^2 t\,dt} . Let \lambda be the value of these integrals, so

    2\lambda  = \int_0^{\pi /2} {dt}  \Rightarrow \lambda  = \frac{\pi }<br />
{4}, and the conclusion follows.
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  4. #4
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    Hi thx for feedback final problem I had was..

    Limits [1,10] (1*x^-1/2)*log(x^1/2) dx where x = t^2
    First thought that came to mind was the square roots would cancel when substituing t and dt = 1/2*t^3

    but the log part left me fogged out only idea I can think of is "(log*x^1/2) ---> 1/t + Constant.

    Anything else I could do?
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