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Thread: keep getting slightly wrong answer for integration question

  1. #1
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    Question keep getting slightly wrong answer for integration question

    "Use the substitution x=secθ where 0<=θ<π/2 to evaluate
    (integrand) √(x^2-1)/x^4 dx."

    so after substitution:
    ⌠ 2
    ⎮ √(SEC(θ) - 1)
    ⎮ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ dx (it still should be dx right? not dθ?)
    ⎮ 4
    ⌡ SEC(θ)

    after using trig identities I get (sigma)(1-sin^2θ)sinθcosθdx
    the I let u = sinθ and -du=cosθdx

    so
    ⌠ 3
    - ⌡ (u - u ) du
    and that is -(u^2)/2+(u^4)/4 +C
    I suspect this is where I mess things up.

    and then I plug back in the value of u to get
    -(sin^2θ)/2+(sin^4θ)/4+c

    my CAS gives me SIN(θ)^2∑(SIN(θ)^2 - 2)/4 but I don't think that's the same
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  2. #2
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    Quote Originally Posted by superdude View Post
    "Use the substitution x=secθ where 0<=θ<π/2 to evaluate
    (integrand) √(x^2-1)/x^4 dx."

    so after substitution:
    ⌠ 2
    ⎮ √(SEC(θ) - 1)
    ⎮ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ dx (it still should be dx right? not dθ?) (**) you need to switch the dx over
    ⎮ 4
    ⌡ SEC(θ)

    after using trig identities I get (sigma)(1-sin^2θ)sinθcosθdx
    the I let u = sinθ and -du=cosθdx

    so
    ⌠ 3
    - ⌡ (u - u ) du
    and that is -(u^2)/2+(u^4)/4 +C
    I suspect this is where I mess things up.

    and then I plug back in the value of u to get
    -(sin^2θ)/2+(sin^4θ)/4+c

    my CAS gives me SIN(θ)^2∑(SIN(θ)^2 - 2)/4 but I don't think that's the same
    With your substitution $\displaystyle x = \sec \theta $then $\displaystyle \int \frac{\sqrt{x^2-1}}{x^4}\, dx = \int \frac{\sqrt{\sec ^2 \theta -1}}{\sec ^4 \theta }\, \sec \theta \tan \theta d\theta = \int \frac{ \tan^2 \theta}{\sec ^3 \theta}\, d \theta = \int \sin^2 \theta \cos \theta\, d \theta$. You should be able to get the rest.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by superdude View Post
    "Use the substitution x=secθ where 0<=θ<π/2 to evaluate
    (integrand) √(x^2-1)/x^4 dx."

    so after substitution:
    ⌠ 2
    ⎮ √(SEC(θ) - 1)
    ⎮ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ dx (it still should be dx right? not dθ?)
    ⎮ 4
    ⌡ SEC(θ)

    after using trig identities I get (sigma)(1-sin^2θ)sinθcosθdx
    the I let u = sinθ and -du=cosθdx

    so
    ⌠ 3
    - ⌡ (u - u ) du
    and that is -(u^2)/2+(u^4)/4 +C
    I suspect this is where I mess things up.

    and then I plug back in the value of u to get
    -(sin^2θ)/2+(sin^4θ)/4+c

    my CAS gives me SIN(θ)^2∑(SIN(θ)^2 - 2)/4 but I don't think that's the same
    When you make the substitution $\displaystyle x=\sec\theta$, it follows that $\displaystyle \,dx=\sec\theta\tan\theta\,d\theta$.

    Thus, the integral becomes $\displaystyle \int\frac{\sqrt{\sec^2\theta-1}\sec\theta\tan\theta\,d\theta}{\sec^4\theta}=\in t\frac{\tan^2\theta}{\sec^3\theta}\,d\theta=\int\s in^2\theta\cos\theta\,d\theta$

    Now let $\displaystyle u=\sin\theta\implies\,du=\cos\theta\,d\theta$

    The integral now becomes $\displaystyle \int u^2\,du=\tfrac{1}{3}u^3+C$

    Back-substituting, we get $\displaystyle \tfrac{1}{3}\sin^3\theta+C$

    Now since $\displaystyle x=\sec\theta$ and $\displaystyle 0\leqslant\theta\leqslant\frac{\pi}{2}$, it follows that $\displaystyle \sin\theta=\frac{\sqrt{x^2-1}}{x}$ (verify).

    Therefore, $\displaystyle \int\frac{\sqrt{x^2-1}}{x^4}\,dx=\frac{\left(x^2-1\right)\sqrt{x^2-1}}{3x^3}+C$.

    Does this make sense?
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by superdude View Post
    "Use the substitution x=secθ where 0<=θ<π/2 to evaluate
    (integrand) √(x^2-1)/x^4 dx."
    $\displaystyle \int\frac{\sqrt{x^2-1}}{x^4}dx$

    let $\displaystyle x=sec\theta$
    $\displaystyle dx=sec\theta(tan\theta)d\theta$

    $\displaystyle \int\left(\frac{\sqrt{sec^2\theta-1}}{sec^4(\theta)}\right)sec(\theta)tan(\theta)d\t heta$

    $\displaystyle \int\left(\frac{\sqrt{tan^2(\theta)}}{sec^3(\theta )}\right)tan(\theta)d\theta$

    $\displaystyle \int\left(\frac{tan^2(\theta)}{sec^3(\theta)}\righ t)d\theta$

    $\displaystyle \int\left(\frac{sec^2(\theta)-1}{sec^3(\theta)}\right)d\theta$

    $\displaystyle \int\left(\frac{sec^2(\theta)}{sec^3(\theta)}\righ t)d\theta-\int\frac{1}{sec^3(\theta)}d\theta$

    I leave the rest for you ...best wishes

    I am always late I start to answer then I see a lot of people post faster than me that's wonderful it is the best forum the questions solved sooooooo quickly I love this thanks to all of you
    Last edited by Krizalid; Jun 3rd 2009 at 12:44 PM.
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  5. #5
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    put $\displaystyle x=\frac1u$ and you'll get an easy integral to solve.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    put $\displaystyle x=\frac1u$ and you'll get an easy integral to solve.
    I was waiting for you to come up with one of your slick substitutions
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  7. #7
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    one thing I don't get

    Quote Originally Posted by Chris L T521 View Post
    Now since $\displaystyle x=\sec\theta$ and $\displaystyle 0\leqslant\theta\leqslant\frac{\pi}{2}$, it follows that $\displaystyle \sin\theta=\frac{\sqrt{x^2-1}}{x}$ (verify).

    Therefore, $\displaystyle \int\frac{\sqrt{x^2-1}}{x^4}\,dx=\frac{\left(x^2-1\right)\sqrt{x^2-1}}{3x^3}+C$.

    Does this make sense?
    I follow up to the point where I get (sin^3θ/3)+C .I don't understand where you get the $\displaystyle x=\sec\theta$ from, is that an identity? when I plug that in for the value of sinθ I still get a differnt answer
    i.e. (√(x^2-1))^3/3+C ≠ $\displaystyle \int\frac{\sqrt{x^2-1}}{x^4}\,dx=\frac{\left(x^2-1\right)\sqrt{x^2-1}}{3x^3}+C$
    Last edited by superdude; Jun 4th 2009 at 12:23 PM. Reason: more concise
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  8. #8
    MHF Contributor Amer's Avatar
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    from this you have $\displaystyle sin\theta=\frac{\sqrt{x^2-1}}{x}$ right

    then from the picture below you can get sec(theta)

    keep getting slightly wrong answer for integration question-sol.jpg
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  9. #9
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    thanks. i get it
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