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Thread: Integration questions

  1. #1
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    Integration questions

    Hi
    Three questions i am having trouble integrating:
    1) $\displaystyle \int{\frac{cosx}{sinx}} dx$
    This is what i have done:
    $\displaystyle \int{\frac{1}{tanx}}$
    so i got:
    $\displaystyle sec^2x + c$

    answers says its: $\displaystyle ln(sinx)+c$

    Someone tell me why is my one incorrect?

    2) $\displaystyle \int{\frac{x+1}{x^2+2x}} dx$
    Need help how to start this.

    3)$\displaystyle \int{\frac{1}{\sqrt{1-9x^2}}}$
    This is what i got: $\displaystyle sin^{-1}3x+c$

    And this what the answers says: $\displaystyle \frac{1}{3}sin^{-1}3x+c$

    Someone tell me why is my one incorrect?

    P.S
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  2. #2
    o_O
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    #1: How did you get $\displaystyle \sec^2 x$ ? Use the substitution: $\displaystyle u = \sin x \ \Rightarrow \ du = \cos x \ dx$ to get a standard integral

    #2: Use the substitution $\displaystyle u = x^2 + 2x \ \Rightarrow \ du = 2(x+1)dx$ to get another standard integral

    #3: How did you do this question? It looks like you simply used a formula. So if you have a table of integrals, you should see that: $\displaystyle \int \frac{1}{\sqrt{1 - (ax)^2} }\ dx \ = \ \frac{1}{a} \arcsin (ax) + C$
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  3. #3
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    Quote Originally Posted by o_O View Post
    #1: How did you get $\displaystyle \sec^2 x$ ? Use the substitution: $\displaystyle u = \sin x \ \Rightarrow \ du = \cos x \ dx$ to get a standard integral
    ok i'll try that.

    Quote Originally Posted by o_O View Post
    #2: Use the substitution $\displaystyle u = x^2 + 2x \ \Rightarrow \ du = 2(x+1)dx$ to get another standard integral
    i'll try that.

    Quote Originally Posted by o_O View Post
    #3: How did you do this question? It looks like you simply used a formula. So if you have a table of integrals, you should see that: $\displaystyle \int \frac{1}{\sqrt{1 - (ax)^2} }\ dx \ = \ \frac{1}{a} \arcsin (ax) + C$
    ok i understand now.
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    3)$\displaystyle \int{\frac{1}{\sqrt{1-9x^2}}}$
    This is what i got: $\displaystyle sin^{-1}3x+c$

    And this what the answers says: $\displaystyle \frac{1}{3}sin^{-1}3x+c$
    If you don't want to use a formula, then you can use trigonometric substitution.

    Notice that

    $\displaystyle \int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}$

    $\displaystyle = \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}$

    $\displaystyle = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}$.


    Now make the substitution $\displaystyle x = \frac{1}{3}\sin{\theta}$ so that $\displaystyle dx = \frac{1}{3}\cos{\theta}\,d\theta$.

    Also notice that $\displaystyle \sin{\theta} = 3x$ and so $\displaystyle \theta = \arcsin{3x}$.


    Then the integral becomes

    $\displaystyle = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1} {3}\cos{\theta}\,d\theta}$

    $\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}$

    $\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9}(1 - \sin^2{\theta})}}\,d\theta}$

    $\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq rt{\cos^2{\theta}}}\,d\theta}$

    $\displaystyle = \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\ ,d\theta}$

    $\displaystyle = \frac{1}{3}\int{1\,d\theta}$

    $\displaystyle = \frac{1}{3}\theta + C$

    $\displaystyle = \frac{1}{3}\arcsin{3x} + C$.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    If you don't want to use a formula, then you can use trigonometric substitution.

    Notice that

    $\displaystyle \int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}$

    $\displaystyle = \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}$

    $\displaystyle = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}$.


    Now make the substitution $\displaystyle x = \frac{1}{3}\sin{\theta}$ so that $\displaystyle dx = \frac{1}{3}\cos{\theta}\,d\theta$.

    Also notice that $\displaystyle \sin{\theta} = 3x$ and so $\displaystyle \theta = \arcsin{3x}$.


    Then the integral becomes

    $\displaystyle = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1} {3}\cos{\theta}\,d\theta}$

    $\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}$

    $\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9}(1 - \sin^2{\theta})}}\,d\theta}$

    $\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq rt{\cos^2{\theta}}}\,d\theta}$

    $\displaystyle = \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\ ,d\theta}$

    $\displaystyle = \frac{1}{3}\int{1\,d\theta}$

    $\displaystyle = \frac{1}{3}\theta + C$

    $\displaystyle = \frac{1}{3}\arcsin{3x} + C$.
    using the u-substitution looks like i longer way of working out it.
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  6. #6
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    This is what i done in the second question. Can you tell me where i have gone wrong, note this is my first time doing u-substitution.

    $\displaystyle u=sinx$

    $\displaystyle du=cosx$

    $\displaystyle \int{\frac{cosx}{u}}$

    $\displaystyle cosx\int{u}$

    $\displaystyle =cosx \frac{u^2}{2} +c$

    $\displaystyle =\frac{cos(x)u^2}{2}$

    substitute $\displaystyle sinx $back into u$\displaystyle

    =\frac{cosxsinx^2}{2}$
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  7. #7
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    You can't factor a variable term like $\displaystyle \cos{x}$ from an integral. Instead of factoring it from the integral, replace it with $\displaystyle du$ because, as you stated, $\displaystyle du = \cos{x}$. Then, use the integration rule $\displaystyle \int \frac{du}{u} = \ln|u| + C$.
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  8. #8
    Ted
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    Quote Originally Posted by Paymemoney View Post
    This is what i done in the second question. Can you tell me where i have gone wrong, note this is my first time doing u-substitution.

    $\displaystyle u=sinx$

    $\displaystyle du=cosx$

    $\displaystyle \int{\frac{cosx}{u}}$

    $\displaystyle cosx\int{u}$

    $\displaystyle =cosx \frac{u^2}{2} +c$

    $\displaystyle =\frac{cos(x)u^2}{2}$

    substitute $\displaystyle sinx $back into u$\displaystyle

    =\frac{cosxsinx^2}{2}$
    For $\displaystyle \int \frac{cos(x)}{sin(x)} dx$ :

    substitute $\displaystyle u=sin(x)$ to get $\displaystyle du=cos(x)dx$

    the integral becomes: $\displaystyle \int \frac{du}{u}=ln|u|+C=ln|sin(x)|+C$

    you should review the substitution method in your book.


    T
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  9. #9
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    ok thanks everyone for the help i understand how to do it now.
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  10. #10
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    Quote Originally Posted by Paymemoney View Post
    using the u-substitution looks like i longer way of working out it.
    I'm just showing you how the standard integral was found in the first place - i.e. through trigonometric substitution.
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