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Math Help - Evaluate the definite Integral

  1. #1
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    Evaluate the definite Integral

    Question : Evaluate

    <br />
\int_{-1}^{1} \ \frac{3 + sin^3 x}{1 + x^2} \ dx<br />
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by zorro View Post
    Question : Evaluate

    <br />
\int_{-1}^{1} \ \frac{3 + sin^3 x}{1 + x^2} \ dx<br />
    <br />
\int_{-1}^{1} \frac{3 + \sin^3 x}{1 + x^2} \ \mathrm{d}x=3\int_{-1}^{1} \frac{1}{1 + x^2} \ \mathrm{d}x+\int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x<br />
    Hint : note that x\mapsto \frac{\sin^3 x}{1 + x^2} is odd.
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  3. #3
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    Stuck again !!!

    Quote Originally Posted by flyingsquirrel View Post
    <br />
\int_{-1}^{1} \frac{3 + \sin^3 x}{1 + x^2} \ \mathrm{d}x=3\int_{-1}^{1} \frac{1}{1 + x^2} \ \mathrm{d}x+\int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x<br />
    Hint : note that x\mapsto \frac{\sin^3 x}{1 + x^2} is odd.

    I am able to integrate the first term but not the last one


    = 3 \int_{-1}^{1} \frac{1}{1 + x^2} + \int_{-1}^{1} \frac{sin^3 x}{1 + x^2}

    = I_1 + I_2



    I_1 = 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1

    = 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1

    = 3 \left[ \frac{\pi}{4} +\frac{\pi}{4} \right] + c_1

    = 3 \left[ \frac{2 \pi}{4}  \right] + c_1

    =  \left[ \frac{6 \pi}{4}  \right] + c_1



    I_2 =  \int_{-1}^{1} \frac{sin^3 x}{1 + x^2} dx_2

    =  \int_{-1}^{1} \frac{sin^3 x}{1 + x^2} dx_2


    I am stuck here!!!!!......
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  4. #4
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    Quote Originally Posted by flyingsquirrel View Post
    Hint : note that x\mapsto \frac{\sin^3 x}{1 + x^2} is odd.
    so \int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x<br />
 = 0
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by zorro View Post
    I_1 = 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1
    There is no need to add a constant since \int_{-1}^1\frac{1}{1+x^2}\,\mathrm{d}x is a definite integral.
    Quote Originally Posted by zorro View Post
    =  \left[ \frac{6 \pi}{4}  \right] + c_1
    \frac{6\pi}{4}=\frac{3\times 2\pi}{2\times 2}=\frac{3\pi}{2}
    Quote Originally Posted by zorro View Post
    I_2 =  \int_{-1}^{1} \frac{sin^3 x}{1 + x^2} dx_2
    I_2 = \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x + \int_0^1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x
    Substitute x=-t ( \implies t=-x and \mathrm{d}t=-\mathrm{d}x) in the first integral :
    \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x =-\int_{1}^0\frac{\sin^3(-t)}{1+(-t)^2}\,\mathrm{d}t=\int_0^1\frac{-\sin^3t}{1+t^2}\,\mathrm{d}t=-\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t
    Thus
    I_2 = -\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t+\int_0^  1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x=0.

    This is a special case of a more general result: Let f:\mathbb{R}\to\mathbb{R} be an odd function and let A be a real number. Then
    \int_{-A}^Af(t)\,\mathrm{d}t=0.

    Edit :
    Quote Originally Posted by dedust View Post
    so \int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x<br />
 = 0
    Yes !
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  6. #6
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    Thanks u very much

    Thank u all very much for helping me
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  7. #7
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    as for that property about the odd function, we also require that f must be integrable there.
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  8. #8
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    what do u mean by that

    Quote Originally Posted by Krizalid View Post
    as for that property about the odd function, we also require that f must be integrable there.

    could u please explain it in non mathematical terms ,as i m not that bright??
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  9. #9
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    Got it

    Quote Originally Posted by zorro View Post
    could u please explain it in non mathematical terms ,as i m not that bright??

    Thanks u all for helping
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