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Math Help - ∫sin3x/sin2x dx

  1. #1
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    ∫sin3x/sin2x dx

    Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.∫sin3x/sin2x dx-minteg.jpg
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    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by gillyjay View Post
    Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.Click image for larger version. 

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    \displaystyle \frac{sin 3x}{sin 2x}=\frac{3sin x-4sin^3 x}{2sin xcos x}=\frac{3-4sin^2 x}{2cos x}=\frac{3-4(1-cos^2x)}{2cosx}=\frac{4cos^2x-1}{2cosx}=2cosx-\frac{secx}{2}

    Can you finish it?
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    Quote Originally Posted by gillyjay View Post
    Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.Click image for larger version. 

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    Numerator: sin(3x) = sin(2x) cos(x) + sin(x) cos(2x).

    Denominator: sin(2x) = 2 sin(x) cos(x).

    Substitute the above, break up the integral into two parts and simplify. The following will be useful: integrate Cos[2x]/Cos[x] - Wolfram|Alpha

    Clcik on Show steps.
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    A probably somewhat an overkill substitution would be to let:
    x = 2\arctan{t} then find  -\int \frac{(t^2-3)(3t^2-1)}{(t-1)(t+1)(t^2+1)^2}\;{dt}.
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    Quote Originally Posted by gillyjay View Post
    Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.Click image for larger version. 

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    \displaystyle\int{\frac{Sin3x}{Sin2x}}dx=\int{\fra  c{Sin2xCosx+Cos2xSinx}{Sin2x}}dx=\int{Cosx}dx+\int  {\frac{Cos2xSinx}{2CosxSinx}dx

    =\displaystyle\int{Cosx}dx+\int{\frac{2Cos^2x-1}{2Cosx}}dx=\int{Cosx}dx+\int{Cosx}dx-\frac{1}{2}\int{Secx}dx
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    Thanks for the hints guys, just a Q, archie, How did you evaluate Sin3x into Sin2xCosx+Cos2xSinx?

    Edit: This problem is indeed eating my mind, how do we go from here to get and to evaluate to get integral value of: 2 sinx + 1/4 ln[1-sinx] / [1 + sinx ] + c (The Right answer)

    Thanks for your answers
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by gillyjay View Post
    Thanks for the hints guys, just a Q, archie, How did you evaluate Sin3x into Sin2xCosx+Cos2xSinx?
    Apply the identity \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\  sin\beta with \alpha=2x and \beta=x (or the other way around...you'll still get the same result).
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    Thanks alot!, hmm but I don't really get the last part on how to get the answer
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    Quote Originally Posted by gillyjay View Post
    Thanks alot!, hmm but I don't really get the last part on how to get the answer
    Do you mean, on arriving at

    \displaystyle\ 2\int{Cosx}dx-\frac{1}{2}\int{Secx}dx

    at which point

    \displaystyle\int{Cosx}dx=Sinx+C

    and

    \displaystyle\int{Secx}dx=ln|Secx+Tanx|+C

    or before that ?

    The other identity used in my previous post was

    \displaystyle\ Cos^2x=\frac{1}{2}\left(1+Cos2x\right)\Rightarrow\ 2Cos^2x-1=Cos2x
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  10. #10
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    If you're wondering how to evaluate \displaystyle \int{\sec{x}\,dx}...


    \displaystyle \int{\sec{x}\,dx} = \int{\frac{1}{\cos{x}}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\cos^2{x}}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{1 - \sin^2{x}}\,dx}

    \displaystyle = \int{\frac{1}{1 - u^2}\,du} after making the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx

    \displaystyle = \int{\frac{1}{(1 - u)(1 + u)}\,du}

    \displaystyle = \int{\frac{1}{2(1 - u)} + \frac{1}{2(1 + u)}\,du}

    \displaystyle = -\frac{1}{2}\ln{|1 - u|} + \frac{1}{2}\ln{|1 + u|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{1 + u}{1 - u}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{1 + \sin{x}}{1 - \sin{x}}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{(1 - \sin{x})(1 + \sin{x})}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{1 - \sin^2{x}}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{\cos^2{x}}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\left(\frac{1 + \sin{x}}{\cos{x}}\right)^2\right|} + C

    \displaystyle = \ln{\left|\left[\left(\frac{1+\sin{x}}{\cos{x}}\right)^2\right]^{\frac{1}{2}}\right|} +C

    \displaystyle = \ln{\left|\frac{1 + \sin{x}}{\cos{x}}\right|} + C

    \displaystyle = \ln{\left|\frac{1}{\cos{x}} + \frac{\sin{x}}{\cos{x}}\right|} + C

    \displaystyle = \ln{|\sec{x} + \tan{x}|} + C.
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    Quote Originally Posted by gillyjay View Post

    This problem is indeed eating my mind, how do we go from here to get and to evaluate to get integral value of: 2 sinx + 1/4 ln[1-sinx] / [1 + sinx ] + c (The Right answer)

    Thanks for your answers
    Also notice that

    \displaystyle\ ln{\left|secx+tanx\right|}=ln{\left|\frac{1}{cosx}  +\frac{sinx}{cosx}\right|}=ln{\left|\frac{1+sinx}{  cosx}\right|}

    =\displaystyle\frac{1}{2}ln\left[\frac{1+sinx}{cosx}\right]^2=\frac{1}{2}ln\left[\frac{(1+sinx)^2}{cos^2x}\right]

    =\displaystyle\frac{1}{2}ln\left[\frac{(1+sinx)(1+sinx)}{1-sin^2x}\right]=\frac{1}{2}ln\left[\frac{(1+sinx)(1+sinx)}{(1-sinx)(1+sinx)}\right]

    =\displaystyle\frac{1}{2}ln{\left|\frac{1+sinx}{1-sinx}\right|}=-\frac{1}{2}ln{\left|\frac{1-sinx}{1+sinx}\right|}


    Multiplying that by a half gives the result in the form you are looking for.
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  12. #12
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    Quote Originally Posted by gillyjay View Post
    Thanks alot!, hmm but I don't really get the last part on how to get the answer
    Please show what you have done and say where you get stuck.
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