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Math Help - trig substitutions

  1. #1
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    trig substitutions

    i cant figure out how to simplify these...please help me out..

    1. integral (sqrt (x^2-a^2))/x^4 dx

    i can get up to integral (a tan^2 x)/(sec^3 x), but i dont know what to do next, because it doesnt easily simplfy...

    2. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx

    3. integral (du)/(u(sqrt(5-u^2)))

    4 integral (dt)/(sqrt(t^2 -6t +13))

    for this one, i think we are supposed to complete the square and then evaluate the integral..
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by skabani View Post
    i cant figure out how to simplify these...please help me out..

    1. integral (sqrt (x^2-a^2))/x^4 dx

    i can get up to integral (a tan^2 x)/(sec^3 x), but i dont know what to do next, because it doesnt easily simplfy...
    i got \frac 1{a^2} \int \frac {\tan^2 \theta}{\sec^3 \theta}~d \theta (check this, i did it in my head )

    if that's true, by using \tan \theta = \frac {\sin \theta}{\cos \theta} and \sec \theta = \frac 1{\cos \theta}, this simplifies to \frac 1{a^2} \int \sin^2 \theta \cos \theta ~d \theta

    now do a substitution. let u = \sin \theta and the rest should be simple


    2. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx
    substitution, u = x^2 + 4

    3. integral (du)/(u(sqrt(5-u^2)))
    you can do a trig substitution here, u = \sqrt{5} \sin \theta

    4 integral (dt)/(sqrt(t^2 -6t +13))

    for this one, i think we are supposed to complete the square and then evaluate the integral..
    yes, that can work. this becomes \int \frac {dt}{\sqrt{(t - 3)^2  + 4}}

    now do a trig substitution. t - 3 = 2 \tan \theta
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  3. #3
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    Hello, skabani!

    1)\;\;\int \frac{\sqrt{x^2-a^2}}{x^4}\,dx

    Let: x \:=\:a\sec\theta\quad\Rightarrow\quad dx \:=\:a\sec\theta\tan\theta\,d\theta

    Substitute: . \int\frac{(a\tan\theta)\,(a\sec\theta\tan\theta\,d  \theta)}{a^4\sec^4\!\theta} \;=\;\frac{1}{a^2}\int\frac{\tan^2\!\theta}{\sec^3  \!\theta}\,d\theta

    . . = \;\frac{1}{a^2}\int \frac{\frac{\sin^2\!\theta}{\cos^2\!\theta}} {\frac{1}{\cos^3\!\theta}}\,d\theta  \;=\;\frac{1}{a^2}\int\sin^2\!\theta\cos\theta\,d\  theta

    Let: u = \sin\theta\quad\Rightarrow\quad du \:=\:\cos\theta\,d\theta

    Substitute: . \frac{1}{a^2}\int u^2\,du \;=\;\frac{1}{3a^2}u^3 + C


    Back-substitute: . \frac{1}{3a^2}\sin^3\!\theta + C


    Back-substitute:

    We have: . \sec\theta \:=\:\frac{x}{a} \:=\:\frac{hyp}{adj} \quad\Rightarrow\quad opp = \sqrt{x^2-a^2} \quad\Rightarrow\quad \sin\theta \:=\:\frac{\sqrt{x^2-a^2}}{x}

    And we have: . \frac{1}{3a^2}\left(\frac{\sqrt{x^2-a^2}}{x}\right)^3 + C \;=\;\frac{1}{3a^2}\cdot\frac{(x^2-a^2)^{\frac{3}{2}}}{x^3} + C

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  4. #4
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    thanks so much!!

    did you happen to get any of the other ones, im having trouble with them too..

    thanks again!!
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    i got \frac 1{a^2} \int \frac {\tan^2 \theta}{\sec^3 \theta}~d \theta (check this, i did it in my head )

    if that's true, by using \tan \theta = \frac {\sin \theta}{\cos \theta} and \sec \theta = \frac 1{\cos \theta}, this simplifies to \frac 1{a^2} \int \sin^2 \theta \cos \theta ~d \theta

    now do a substitution. let u = \sin \theta and the rest should be simple

    substitution, u = x^2 + 4

    you can do a trig substitution here, u = \sqrt{5} \sin \theta

    yes, that can work. this becomes \int \frac {dt}{\sqrt{(t - 3)^2 + 4}}

    now do a trig substitution. t - 3 = 2 \tan \theta

    thanks...so for 2, i got 1/8 - 1/4sqrt(5) + c
    3. 1/5 ln u/sqrt(5) ln (sqrt(5) + u)/sqrt(5-^2)) + c
    4. 1/3 arctan (x-3)/2 + c

    are these right?
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