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Thread: indefinite integral of 4cos+sinx-cosx/ sin^2x

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    indefinite integral of 4cos+sinx-cosx/ sin^2x

    4cos+sinx-cosx/ sin^2x
    How do I start? I don't think I can use u substitution:/
    I tried canceling the sinx bit I'm still left with sinx at the bottom.
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    Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

    Hello Dream,

    We have the following:
     \int \frac{4cosx+sinx-cosx}{sin^2x}dx \implies \int \frac{3cosx+sinx}{sin^2x}dx

    =3\int \frac{cosx}{sin^2x}dx+ \int \frac{1}{sinx}dx

    Can you finish it?


    [hint] For the first substitution, let u = sinx , for the second write  \frac{1}{sinx}=cscx [/hint]
    Last edited by sakonpure6; May 8th 2015 at 05:29 PM.
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  3. #3
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    Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

    Quote Originally Posted by dreamd7358 View Post
    4cos+sinx-cosx/ sin^2x
    How do I start? I don't think I can use u substitution:/
    I tried canceling the sinx bit I'm still left with sinx at the bottom.
    well, what if the integrand is as follows?

    \int 4\cos{x} + \sin{x} - \frac{\cos{x}}{\sin^2{x}} \, dx =

    \int 4\cos{x} + \sin{x} - \csc{x}\cot{x} \, dx = 4\sin{x} - \cos{x} + \csc{x} + C
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    Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

    Hello, dreamd7358!

    \int \frac{4\cos x+\sin x-\cos x}{\sin^\!}\,dx

    We have: . \int\frac{3\cos x + \sin x}{\sin^2\!x}\,dx \;=\;\int\left(\frac{\3\cos x}{\sin^2\!x} + \frac{\sin x}{\sin^2\!x}\right)dx

    . . . . . . =\;\int\left(3\!\cdot\!\frac{1}{\sin x}\!\cdot\!\frac{\cos x}{\sin x} + \frac{1}{\sin x}\right)dx \;=\;\int\left(3\csc x\cot x + \csc x)\,dx

    . . . . . . =\;-3\csc x + \ln\big|\csc x - \cot x\big|+ C
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  5. #5
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    Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

    Notice that skeeter did a completely different problem from sakonpure6 and Soroban! dreamd7358, please use parentheses to make you meaning clear. skeeter assumed you meant what you actually wrote: 4cos+sinx-cosx/ sin^2x= 4cos(x)+ sin(x)- (cos(x)/sin^2(x)) while sakonpure6 and Soroban guess that you really meant (4 cos(x)+ sin(x)- cos(x))/sin^2(x).

    (Althoug we all had to guess that your initial "4cos+ sinx" meant "4cos(x)+ sin(x)"!
    Thanks from sakonpure6
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