calculute [(sin x)^1/2]dx?

  1. profile image47
    karzan.ahmadposted 7 years ago

    calculute [(sin x)^1/2]dx?

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    bootan83posted 7 years ago

    your quation is wronge please insert the integral , but the quation is not answer exactly, in numerical u can get approximation solution.

  3. Ben Evans profile image73
    Ben Evansposted 7 years ago

    If this is the integral of [(sin x)^1/2]dx this solves into another elliptical integral.  Lets just suffice to say that it is not easily solved.

    Why?

    It cannot be U substituted.  A integrand of say (sinx)^1/2(cosx)dx is going to produce a du equal to -cosxdx but our original function doesn't.

    Functions like Int[(sin x)^3]dx  becomes Int[1-cos^2(x)]sinxdx.  Again without too much ado du=sinxdx these are called odd functions.

    Functions like Int[(sinx)^2dx} can be solved with half angle identities:
    (sinx)^2=1/2(1-cos2x)

    Okay we cant do integration by parts because terms do not drop out.

    If you do U substitution you are going to end up with this

    Int[u^1/2/(1-u^2)^1/2]du

    The denominator of the integrand cannot be "u" (we would define it v) substituted.  Further more this cannot be trig substituted because it will give us the same thing we started with.  It cannot be algebraically manipulated.  Partial fraction produce a form A(1-u) +B(1+u) =u^1/2.

    This does solve into a very complex function.....Heck it is another integral woohoo. Take a look around and see what you see.  There are no simple solutions and there is a reason.

 
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