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Math Help - Integrations with Trigonometric Substitutions.

  1. #1
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    Integrations with Trigonometric Substitutions.

    I have a few variations of some problems that would make the entire homework a lot easier. However, our teacher never covered some of these problems, and I would appreciate some assistance?

    a. Integral of 1/(x^2(x^2 - 1)^(1/2)) dx

    b. Integral of (x^3)/((x^2 + 4)^(1/2)) dx

    c. Integral of ((9 - x^2)^(1/2))/x^2 dx

    This is all under Trigonometric substitutions, where substitutions include those sec, cos, and tan.

    :S

    For a., that extra x squared is miffing me. B. the variable in the numerator prevents a convenient substitution. C. x squared in the denominator is unlike anything we were taught in class.

    Help?
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  2. #2
    Senior Member DivideBy0's Avatar
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    Hi! niyati, while waiting for someone more competent to help you, here are some tips:

    Usually if the integral contains \sqrt{a^2-x^2}, where a is constant, you can use the substitution x=a\sin{\theta}

    When the integral contains \sqrt{a^2+x^2}, try x=a\tan{\theta}.

    When the integral contains \sqrt{x^2-a^2} try x=a\sec{\theta}

    It's got to do with the common trigonometric identities. Work through a few problems and you'll see it simplifies very nicely. Trust me.
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  3. #3
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    Integration by trig substitution.
    Sqrt of something.

    Umm, the Pythagorean trig identity,
    sin^2(A) +cos^2(A) = 1 ----------(i)
    Divide both sides by cos^2(A),
    tan^2(A) +1 = sec^2(A) ----------(ii)

    a. Integral of 1/(x^2(x^2 - 1)^(1/2)) dx

    The "sqrt something" is sqrt[x^2 -1].
    So we use (ii), or variations of it,
    tan^2(A) = sec^2(A) -1 ----------------(iii)
    tanA = sqrt[sec^2(A) -1] ------------------------***

    Let x = secU -----------***
    So,
    ----x^2 = sec^2(U)
    ----dx = secUtanU dU
    ----sqrt[x^2 -1] = sqrt[sec^2(U) -1] = tanU

    = INT.[1 / (sec^2(U)*tanU)]*(secUtanU dU)
    = INT.[1 /secU]du
    = INT.[cosU]dU
    = sinU +C

    What is sinU?

    x = secU
    sec = hypotenuse/adjacent side
    So, in the reference right triangle of angle u,
    --hypotenuse = x
    --adjacent side = 1
    --opposite side = sqrt[x^2 -1]
    So,
    sinU = opp/hyp = sqrt[x^2 -1] / x

    Therefore,
    = {sqrt[x^2 -1] / x} +C ----------------------answer.
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  4. #4
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    b. Integral of (x^3)/((x^2 + 4)^(1/2)) dx

    The "sqrt something" is sqrt[x^2 +4].
    So we use tan^2(A) +1 = sec^2(A).
    secA = sqrt[tan^2(A) +1] -----------------------***

    sqrt[x^2 +4]
    = sqrt[4(x^2)/4 +4/4]
    = 2sqrt[(x/2)^2 +1]

    Hence, let x/2 = tanU ------------***
    So,
    ---x = 2tanU
    ---x^3 = 8tan^3(U)
    ---dx = 2sec^2(U) dU
    ---sqrt[x^2 +4] = 2sqrt[(x/2)^2 +1] = 2sqrt[tan^2(U) +1] = 2secU

    Then,
    INT.[(x^3)/(sqrt(x^2 + 4))]dx
    = INT.[(8tan^3(U)) / 2secU]*(2sec^2(U) dU)
    = (8)INT.[tan^3(U)secU]dU

    Umm, secU dU is not the derivative of tanU.
    Neither is secUtanU du.
    So, cannot integrate yet.

    = (8)INT.[tan^2(U)]*secUtanU du
    = (8)INT.[sec^2(U) -1]*dsecU
    = 8[(1/3)sec^3(U) -secU] +C

    What is secU?
    x/2 = tanU
    So,
    --opp side = x
    --adj side = 2
    --hypotenuse = sqrt[x^2 +4]
    And so,
    secU = hyp/adj = sqrt[x^2 +4] / 2

    Hence,
    = 8[(1/3){sqrt[x^2 +4] /2}^3 -sqrt[x^2 +4] /2] +C
    = (1/3)[x^2 +4]^(3/2) -4[x^2 +4]^(1/2) +C --------------answer.
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  5. #5
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    c. Integral of ((9 - x^2)^(1/2))/x^2 dx

    The "sqrt something" is sqrt[9 -x^2]
    So we use
    sin^2(A) +cos^2(A) = 1
    or variation pof it.
    cos^2(A) = 1 -sin^2(A) ---------------***

    sqrt[9 -x^2]
    = sqrt[9/9 -(x^2)/9]
    = 3sqrt[1 -(x/3)^2]

    So, let x/3 = sinU ---------------***
    Then,
    --x = 3sinU
    --x^2 = 9sin^2(U)
    --dx = 3cosU
    --sqrt[9 -x^2] = 3sqrt[1 -(x/3)^2] = 3sqrt[1 -sin^2(U)] = 3cosU

    Then,
    INT.[sqrt(9 - x^2) / x^2] dx
    = INT.[3cosU / 9sin^2(U)] (3cosU du)
    = INT.[cos^2(U) / sin^2(U)]du
    = INT.[cot^2(U)]dU
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  6. #6
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    Quote Originally Posted by niyati View Post
    a. Integral of 1/(x^2(x^2 - 1)^(1/2)) dx

    b. Integral of (x^3)/((x^2 + 4)^(1/2)) dx
    Sometimes, we don't need to apply trig. substitutions. Let's see another methods:

    \int\frac1{x^2\sqrt{x^2-1}}\,dx

    The key is defined by the reciprocal substitution x=\frac1u\implies dx=-\frac1{u^2}\,du, after some simple calculations the integral becomes to

    - \int {\frac{u}<br />
{{\sqrt {1 - u^2 } }}\,du} = \int {\frac{{\left( {1 - u^2 } \right)'}}<br />
{{2\sqrt {1 - u^2 } }}\,du} = \sqrt {1 - u^2 } + k

    Now back substitute and yields

    \int\frac1{x^2\sqrt{x^2-1}}\,dx=\frac{\sqrt{x^2-1}}x+k.

    --

    \int\frac{x^3}{\sqrt{x^2+4}}\,dx

    It suffices to set u=\sqrt{x^2+4} (you'll see why.)
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