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Math Help - Reduction Integral Formula (tan)

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    Reduction Integral Formula (tan)

    Q: I_n(x) = \displaystyle\int^x_0 \tan ^x \theta \, \mathrm{d}\theta, \ \ n \ge 0, |x|<\frac{\pi}{2}

    By writing \tan \theta as \tan ^{n-2} \theta \tan ^2 \theta, or otherwise, show that

    I_x(x) = \frac{1}{n-1} \tan ^{n-1}x - I_{n-2}(x), \ \ n \ge 2, |x| <\frac{\pi}{2}

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    Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance.
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    Quote Originally Posted by Air View Post
    Q: I_n(x) = \displaystyle\int^x_0 \tan ^x \theta \, \mathrm{d}\theta, \ \ n \ge 0, |x|<\frac{\pi}{2}

    By writing \tan \theta as \tan ^{n-2} \theta \tan ^2 \theta, or otherwise, show that

    I_x(x) = \frac{1}{n-1} \tan ^{n-1}x - I_{n-2}(x), \ \ n \ge 2, |x| <\frac{\pi}{2}

    __________________
    Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance.
    I_n(x) = \displaystyle\int^x_0 \tan ^n \theta \, \mathrm{d}\theta

    I_n(x) = \displaystyle\int^x_0 \tan^{n-2}\theta  (sec^2 \theta -1) \theta \, \mathrm{d}\theta

    I_n(x) = \int^x_0 \tan^{n-2}\theta  (sec^2 \theta-1)  \, \mathrm{d}\theta

    I_n(x) = \int^x_0 \tan^{n-2}\theta sec^2 \theta \, \mathrm{d}\theta  - \int^x_0 \tan^{n-2}\theta \mathrm{d}\theta

    I_n(x) =  \frac{1}{n-1} \left[ tan^{n-1} \theta \right]^{x}_{0} - I_{n-2}(x)

    I_n(x) =  \frac{1}{n-1} tan^{n-1}x - I_{n-2}(x)


    Bobak

    P.S are you sitting the exam on the 18th of june?
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    Quote Originally Posted by bobak View Post
    I_n(x) = \displaystyle\int^x_0 \tan ^n \theta \, \mathrm{d}\theta

    I_n(x) = \displaystyle\int^x_0 \tan^{n-2}\theta  (sec^2 \theta -1) \theta \, \mathrm{d}\theta

    I_n(x) = \int^x_0 \tan^{n-2}\theta  (sec^2 \theta-1)  \, \mathrm{d}\theta

    I_n(x) = \int^x_0 \tan^{n-2}\theta sec^2 \theta \, \mathrm{d}\theta  - \int^x_0 \tan^{n-2}\theta \mathrm{d}\theta

    I_n(x) =  \frac{1}{n-1} \left[ tan^{n-1} \theta \right]^{x}_{0} - I_{n-2}(x)

    I_n(x) =  \frac{1}{n-1} tan^{n-1}x - I_{n-2}(x)


    Bobak

    P.S are you sitting the exam on the 18th of june?
    Oh, so you didn't need to use integration by parts?

    And, yes, exam on that day.
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  4. #4
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    Quote Originally Posted by Air View Post
    Oh, so you didn't need to use integration by parts?

    And, yes, exam on that day.
    Reduction formula do not always need the use of parts. If you own the heinemann FP2 book look at question 73 on page 126, you don't even need to use any integration method for that.

    \tan^{n-2}\theta sec^2 \theta you should spot as a standard form f^n(x) f'(x) in which cases you could use the subistution u = f(x) or just do it straight off.

    Bobak
    Last edited by bobak; May 8th 2008 at 10:33 AM.
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  5. #5
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    Hello,

    Quote Originally Posted by Air View Post
    Q: I_n(x) = \displaystyle\int^x_0 \tan ^x \theta \, \mathrm{d}\theta, \ \ n \ge 0, |x|<\frac{\pi}{2}

    By writing \tan \theta as \tan ^{n-2} \theta \tan ^2 \theta, or otherwise, show that

    I_x(x) = \frac{1}{n-1} \tan ^{n-1}x - I_{n-2}(x), \ \ n \ge 2, |x| <\frac{\pi}{2}

    __________________
    Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance.
    If you want to integrate by parts and use the given indication... :

    \begin{aligned} I_n(x) &=\int_0^x \tan^n \theta d\theta \\<br />
&=\int_0^x \tan^{n-2} \theta \cdot \tan^2 \theta d \theta \\<br />
&=\int_0^x \tan^{n-2} \theta \cdot (\tan^2 \theta +1)-\tan^{n-2} \theta d \theta \\<br />
&=\int_0^x \tan^{n-2} \theta \cdot (\tan^2 \theta +1) d \theta-I_{n-2}(x)<br />
\end{aligned}

    Now, integrate by parts the first term, knowing that the derivative of tan(x) is 1+tanē(x), and you got it
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