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Math Help - Integrating Trigonometry

  1. #1
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    Johor Bahru
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    Integrating Trigonometry

    I tried doing 2 questions here which I fail to get the answer.

    1) ∫ tan^2 (1-3x) dx

    f(x) = tan(1-3x)
    f'(x) = -3 sec^2 (1-3x)

    By using the integration technique, that is ,
    ∫f(x)^n * f'(x) = f(x)^(n+1) /(n+1) + c

    I'll get,
    (1/ (-3sec^2 (1-3x)) * ( tan(1-3x)^3 / 3).
    I know something is wrong somewhere, because the answer is -1/3 tan(1-3x) -x + c

    2) The same goes to the second question, ∫  sin^2 (2x) dx.
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  2. #2
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    Re: Integrating Trigonometry

    1. Let u = 1-3x so that du = -3dx \text{  and  } dx = -\dfrac{du}{3}

    Hence we have \displaystyle -\dfrac{1}{3} \int \tan^2(u)\, du which may or may not be a standard integral. I'll outline the integration below:

    We know from our trig identities that \tan^2(u) = \sec^2(u)-1 so let's sub that into the integral:

    \int \tan^2(u)\, du = \int (\sec^2(u) - 1)du

    We know that \dfrac{d}{du} \tan(u) = \sec^2(u) so then the reverse must be true: \int \sec^2(u) = \tan(u) (the constant will be added later)

    Thus \int (\sec^2(u)-1)du = \tan(u) - u + C where C is the constant of integration.

    Back substitute to get your integral in terms of x


    2. I'd go with a double angle identity:

    \cos(2\theta) = 1-2\sin^2(\theta) which can be rearranged to give \sin^2(\theta) = \dfrac{1}{2} - \dfrac{\cos(2\theta)}{2}

    In your example \theta = 2x and hence \sin^2(2x) = \dfrac{1}{2} - \dfrac{\cos(4x)}{2}.

    The integral of cos(ax) is \int \cos(ax) dx = \dfrac{1}{a} \sin(ax) for constant, non-zero a.
    Last edited by e^(i*pi); June 24th 2011 at 04:03 AM. Reason: minus sign again
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  3. #3
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    Re: Integrating Trigonometry

    Thanks, I'm able to understand how to get the answer. So, should I substitute trigo identities when I come across similar questions like these?
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    Re: Integrating Trigonometry

    Quote Originally Posted by nav92 View Post
    Thanks, I'm able to understand how to get the answer. So, should I substitute trigo identities when I come across similar questions like these?
    If you can't solve them immediately then often an identity will make it easier to solve. The main one I come across is \cos(2\theta) = 2\cos^2(\theta)-1 = 1-2\sin^2(\theta) because it allows you to change a squared term into a linear term.
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  5. #5
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    Re: Integrating Trigonometry

    You should know that \displaystyle \int{\sec^2{x}\,dx} = \tan{x}. You should also know that \displaystyle 1 + \tan^2{x} \equiv \sec^2{x}. So

    \displaystyle \begin{align*}\int{\tan^2{(1 - 3x)}\,dx} &= \int{\sec^2{(1 - 3x)} - 1} \\ &= -\frac{1}{3}\int{-3\sec^2{(1 - 3x)}\,dx}  - \int{1\,dx} \\ &= -\frac{1}{3}\int{\sec^2{(u)}\,du} - \int{1\,dx}\textrm{ after making the substitution }u = 1 - 3x \implies du = -3\,dx \\ &= -\frac{1}{3}\tan{(u)} - x + C \\ &= -\frac{1}{3}\tan{(1 - 3x)} - x + C \end{align*}
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  6. #6
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    Re: Integrating Trigonometry

    Quote Originally Posted by nav92 View Post
    I tried doing 2 questions here which I fail to get the answer.

    1) ∫ tan^2 (1-3x) dx

    f(x) = tan(1-3x)
    f'(x) = -3 sec^2 (1-3x)

    By using the integration technique, that is ,
    ∫f(x)^n * f'(x) = f(x)^(n+1) /(n+1) + c
    The problem is that you don't have \int f^n f' dx because you don't have the f'.
    You can't just write it as
    \frac{1}{-3sec^2(1- 3x)} \int tan^2(1- 3x)(-3sec^2(1- 3x))dx
    because you cannot pass a function of x through the integral like that.

    I'll get,
    (1/ (-3sec^2 (1-3x)) * ( tan(1-3x)^3 / 3).
    I know something is wrong somewhere, because the answer is -1/3 tan(1-3x) -x + c

    2) The same goes to the second question, ∫  sin^2 (2x) dx.
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