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  1. #1
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    Difficult integration question

    How do I do part (b). I've been stuck at it for hours, even after looking at all the different types of integration formulas...
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    Re: Difficult integration question

    For the first 2=e^x+2-e^x and for the second after a substitution you have to compute, up to a constant, \int \frac{t^2-1}tdt.
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    Re: Difficult integration question

    \int \sqrt{16x^2-9} \cdot \frac{1}{x} \, dx

    3 \int \sqrt{\frac{16x^2}{9} - 1} \cdot \frac{1}{x} \, dx

    x = \frac{3\sec{t}}{4}

    dx = \frac{3}{4} \sec{t}\tan{t} \, dt

    substitute and simplify ...

    3 \int \tan^2{t} \, dt

    3 \int \sec^2{t} - 1 \, dt

    3 \left(\tan{t} - t) + C

    \sqrt{16x^2-9} - 3\arctan\left(\frac{\sqrt{16x^2-9}}{3}\right) + C
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    Re: Difficult integration question

    Quote Originally Posted by skeeter View Post
    \int \sqrt{16x^2-9} \cdot \frac{1}{x} \, dx

    3 \int \sqrt{\frac{16x^2}{9} - 1} \cdot \frac{1}{x} \, dx

    x = \frac{3\sec{t}}{4}

    dx = \frac{3}{4} \sec{t}\tan{t} \, dt

    substitute and simplify ...

    3 \int \tan^2{t} \, dt

    3 \int \sec^2{t} - 1 \, dt

    3 \left(\tan{t} - t) + C

    \sqrt{16x^2-9} - 3\arctan\left(\frac{\sqrt{16x^2-9}}{3}\right) + C
    How do you know we should let x = 3sec t/ 4 ? I've been listening in class and I don't recall my lecturer showing us such a method? Did you use trial and error to get x = 3sect/4?
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    Re: Difficult integration question

    Quote Originally Posted by Flipflops View Post
    How do you know we should let x = 3sec t/ 4 ? I've been listening in class and I don't recall my lecturer showing us such a method? Did you use trial and error to get x = 3sect/4?
    It's a method known as trigonometric substitution, and takes some inspired guesswork. Basically, the fact that derivatives of trig functions are trig functions, and through the Pythagorean Theorem, we can usually simplify a trig expression and cancel with the derivative, to make the integral easy to evaluate.

    Here we have something of the form \displaystyle \begin{align*} a^2x^2 - b^2 \end{align*}, so the identity \displaystyle \begin{align*} \sec^2{\theta} - 1 \equiv \tan^2{\theta} \end{align*} tells us a substitution of the form \displaystyle \begin{align*} x = \frac{b\sec{\theta}}{a} \end{align*} is appropriate.

    \displaystyle \begin{align*} a^2x^2 - b^2 &= a^2 \left( \frac{b\sec{\theta}}{a} \right)^2 - b^2 \\ &= a^2 \left( \frac{b^2\sec^2{\theta}}{a^2} \right) - b^2 \\ &= b^2\sec^2{\theta} - b^2 \\ &= b^2\left(\sec^2{\theta} - 1\right) \\ &= b^2\tan^2{\theta} \end{align*}

    And notice that \displaystyle \begin{align*} \frac{d}{dx}\left(\sec{\theta}\right) &= \sec{\theta}\tan{\theta} \end{align*}, which makes a combination that is easy to evaluate, as per Skeeter's post.
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    Re: Difficult integration question

    Hello, Flipflops!

    There is a clever "trick" for the first one . . .


    \displaystyle(a)\;\int\frac{dx}{e^x+2}

    Divide the numerator and denominator by e^x.

    \text{We have: }\:\int\frac{e^{\text{-}x}\,dx}{1 + 2e^{\text{-}x}}

    \text{Let }\,u \,=\,1 + 2e^{-x} \quad\Rightarrow\quad du \,=\,-2e^{-x}dx \quad\Rightarrow\quad e^{-x}dx \:=\:-\tfrac{1}{2}\:\!du

    \text{Substitute: }\:\int\frac{\text{-}\frac{1}{2}du}{u} \;=\;\text{-}\tfrac{1}{2}\int\frac{du}{u} \;=\;\text{-}\tfrac{1}{2}\ln|u| + C

    \text{Back-substitute: }\:\text{-}\tfrac{1}{2}\ln(1 + 2e^{-x}) + C


    This can be simplified beyond all recognition . . .

    -\tfrac{1}{2}\ln(1 + 2e^{-x}) + C \;=\;-\tfrac{1}{2}\ln\left(1 + \frac{2}{e^x}\right) + C \;=\;-\tfrac{1}{2}\ln\left(\frac{e^x + 2}{e^x}\right) + C

    . . =\;-\tfrac{1}{2}\big[\ln(e^x+2) - \ln(e^x)\big] + C \;=\; -\tfrac{1}{2}\big[\ln(e^x+2) - x\big] + C

    . . =\;\tfrac{1}{2}\big[x - \ln(e^x+2)\big]+C

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    Re: Difficult integration question

    Quote Originally Posted by Soroban View Post
    Hello, Flipflops!

    There is a clever "trick" for the first one . . .



    Divide the numerator and denominator by e^x.

    \text{We have: }\:\int\frac{e^{\text{-}x}\,dx}{1 + 2e^{\text{-}x}}

    \text{Let }\,u \,=\,1 + 2e^{-x} \quad\Rightarrow\quad du \,=\,-2e^{-x}dx \quad\Rightarrow\quad e^{-x}dx \:=\:-\tfrac{1}{2}\:\!du

    \text{Substitute: }\:\int\frac{\text{-}\frac{1}{2}du}{u} \;=\;\text{-}\tfrac{1}{2}\int\frac{du}{u} \;=\;\text{-}\tfrac{1}{2}\ln|u| + C

    \text{Back-substitute: }\:\text{-}\tfrac{1}{2}\ln(1 + 2e^{-x}) + C


    This can be simplified beyond all recognition . . .

    -\tfrac{1}{2}\ln(1 + 2e^{-x}) + C \;=\;-\tfrac{1}{2}\ln\left(1 + \frac{2}{e^x}\right) + C \;=\;-\tfrac{1}{2}\ln\left(\frac{e^x + 2}{e^x}\right) + C

    . . =\;-\tfrac{1}{2}\big[\ln(e^x+2) - \ln(e^x)\big] + C \;=\; -\tfrac{1}{2}\big[\ln(e^x+2) - x\big] + C

    . . =\;\tfrac{1}{2}\big[x - \ln(e^x+2)\big]+C

    Or another alternative is to multiply top and bottom by \displaystyle \begin{align*} e^x \end{align*}.

    \displaystyle \begin{align*} \int{\frac{dx}{e^x + 2}} &= \int{\frac{e^x\,dx}{e^{2x} + 2e^x}} \\ &= \int{\frac{e^x\,dx}{e^{2x} + 2e^x + 1^2 - 1^2}} \\ &= \int{\frac{e^x\,dx}{\left(e^x + 1\right)^2 - 1}} \\ &= \int{\frac{du}{u^2 - 1}}\textrm{ after making the substitution }u = e^x + 1 \\ &= \int{\frac{1}{2(u - 1)} - \frac{1}{2(u + 1)}\,du} \\ &= \frac{1}{2}\ln{|u - 1|} - \frac{1}{2}\ln{|u + 1|} + C \\ &= \frac{1}{2}\ln{\left|\frac{u - 1}{u + 1}\right|} + C \\ &= \frac{1}{2}\ln{\left|\frac{e^x + 1 - 1}{e^x + 1 + 1}\right|} + C \\ &= \frac{1}{2}\ln{\left|\frac{e^x}{e^x + 2}\right|} + C \end{align*}
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    Re: Difficult integration question

    Quote Originally Posted by Flipflops View Post
    How do you know we should let x = 3sec t/ 4 ? I've been listening in class and I don't recall my lecturer showing us such a method? Did you use trial and error to get x = 3sect/4?
    ask your "lecturer" about trig subs ...

    Pauls Online Notes : Calculus II - Trig Substitutions
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