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Math Help - Evaluating the Integrals

  1. #1
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    Evaluating the Integrals

    I've got these two integral problems, and I'm stuck with these.
    Can anybody give me some hints for these two questions?
    Thanks for answering.
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  2. #2
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    \int \csc^4{x}\cot^6{x} \, dx

    \int \csc^2{x}(\cot^2{x}+1)\cot^6{x} \, dx

    -\int (\cot^8{x} + \cot^6{x})(-\csc^2{x}) \, dx

    u = \cot{x}

    du = -\csc^2{x} \, dx

    -\int (u^8 + u^6) \, du
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  3. #3
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    Quote Originally Posted by Gilvanildo View Post
    I've got these two integral problems, and I'm stuck with these.
    Can anybody give me some hints for these two questions?
    Thanks for answering.
    For b), try completing the square in the denominator, then using the substitution x + 1 = 2\sec{\theta}.
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  4. #4
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    Hello, Gilvanildo!

    (a)\;\int \csc^4\!x\cot^6\!x\,dx

    We have: . \csc^4\!x\cot^6\!x \;=\;\csc^2\!x\cot^6\!x\,(\csc^2\!x\,dx)

    . . . . . . . . . . . . . . . = \;(\cot^2\!x+1)\cot^6\!x\,(\csc^2\!x\,dx)

    . . . . . . . . . . . . . . . = \;(\cot^8\!x + \cot^6\!x)\,(\csc^2\!x\,dx)


    Then we have: . \int(\cot^8\!x + \cot^6\!x)\,(\csc^2\!x\,dx)


    Let: . u \,=\,\cot x \quad\Rightarrow\quad dx \,=\,-\csc^2\!x\,dx \quad\Rightarrow\quad \csc^2\!x\,dx \,=\,-du

    . . and substitute . . .




    (b)\;\int\frac{x\,dx}{\sqrt{x^2+2x-3}}
    Under the radical, complete the square: . \sqrt{x^2+2x-3} \:=\:\sqrt{(x+1)^2 - 4}

    Then we have: . \int\frac{x\,dx}{\sqrt{(x+1)^2-4}}


    Let: . x + 1 \:=\:2\sec\theta \quad\Rightarrow \quad dx \:=\:2\sec\theta\tan\theta\,d\theta

    Also: . x \:=\:2\sec\theta - 1 \quad\hdots\quad \sqrt{(x+1)^2-4} \:=\:2\tan\theta


    Substitute: . \int\frac{(2\sec\theta-1)(2\sec\theta\tan\theta\,d\theta)}{2\tan\theta} \;=\; \int(2\sec^2\!\theta - \sec\theta)\,d\theta

    Got it?



    Edit: Too slow again . . . *blush*
    .
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