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Math Help - integration by part and definate integrate

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    Integration by part and definate integrate

    Please help me to check whether the answer is correct or not.
    1. ∫1/x^2 ln x dx = -1/x ln x + 1/x^2 + c
    2. ∫x sin 2x dx = -x cos 2x/2 + sin 2x/4 + c
    3. ∫_3^5 5/(2x-5) dx = 5/2 ln 5
    4. ∫_0^π/4 (e^-2x + 2 tan x) dx = 1/2(1-e^((-π)/2)/2)

    Help me to prove these two.
    1. ∫_0^π/4 tan^3 x dx = (1-ln 2/2)
    2. ∫_0^π/2 4/(3+5sin x) dx = ln 3

    And thanks in advance, I already work on it for few hours, thus your help will be greatly appreciated.
    Last edited by alexander9408; February 25th 2013 at 08:53 AM.
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  2. #2
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    Re: integration by part and definate integrate

    integral - Wolfram|Alpha

    use this to check results

    Greetings
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  3. #3
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    Re: Integration by part and definate integrate

    Quote Originally Posted by alexander9408 View Post
    Please help me to check whether the answer is correct or not.
    1. ∫1/x^2 ln x dx = -1/x ln x + 1/x^2 + c
    2. ∫x sin 2x dx = -x cos 2x/2 + sin 2x/4 + c
    3. ∫_3^5 5/(2x-5) dx = 5/2 ln 5
    4. ∫_0^π/4 (e^-2x + 2 tan x) dx = 1/2(1-e^((-π)/2)/2)

    Help me to prove these two.
    1. ∫_0^π/4 tan^3 x dx = (1-ln 2/2)
    2. ∫_0^π/2 4/(3+5sin x) dx = ln 3

    And thanks in advance, I already work on it for few hours, thus your help will be greatly appreciated.
    1.
    \displaystyle \begin{align*} \int{\frac{\ln{(x)}}{x^2}\,dx} &= \int{x^{-2}\ln{(x)}\,dx} \end{align*}

    Let \displaystyle \begin{align*} u = \ln{(x)} \implies du = x^{-1}\,dx \end{align*} and \displaystyle \begin{align*} dv = x^{-2}\,dx \implies v = -x^{-1} \end{align*} and we find through integrating by parts...

    \displaystyle \begin{align*} \int{x^{-2}\ln{(x)}\,dx} &= -x^{-1}\ln{(x)} - \int{-x^{-2}\,dx} \\ &= -\frac{\ln{(x)}}{x} + \int{x^{-2}\,dx} \\ &= -\frac{\ln{(x)}}{x} - x^{-1} + C \\ &= -\frac{\ln{(x)}}{x} - \frac{1}{x} + C \\ &= -\frac{\left[ 1 + \ln{(x)} \right] }{ x} + C  \end{align*}


    2.
    \displaystyle \begin{align*} \int{x\sin{(2x)}\,dx} &= -\frac{1}{2}x\cos{(2x)} - \int{ -\frac{1}{2}\cos{(2x)}\,dx } \\ &= -\frac{1}{2}x\cos{(2x)} + \frac{1}{2}\int{\cos{(2x)}\,dx} \\ &= -\frac{1}{2}x\cos{(2x)} + \frac{1}{4}\sin{(2x)} + C \end{align*}


    3.
    \displaystyle \begin{align*} \int_3^5{\frac{5}{2x - 5}\,dx} &= \frac{5}{2}\int_3^5{\frac{2}{2x - 5}\,dx} \\ &= \frac{5}{2}\int_1^5{\frac{1}{u}\,du} \textrm{ after making the substitution } u = 2x - 5 \\ &= \frac{5}{2} \left[ \ln{|u|} \right]_1^5 \\ &= \frac{5}{2} \left[ \ln{|5|} - \ln{|1|} \right] \\ &= \frac{5}{2}\ln{(5)} \end{align*}


    4.
    \displaystyle \begin{align*} \int_0^{\frac{\pi}{4}}{e^{-2x} + 2\tan{(x)}\,dx} &= \int_0^{\frac{\pi}{4}}{e^{-2x}\,dx} - 2 \int_0^{\frac{\pi}{4}}{\frac{-\sin{(x)}}{\cos{(x)}}\,dx} \\ &= \left[ -\frac{1}{2}e^{-2x} \right]_0^{\frac{\pi}{4}} - 2\int_1^{\frac{1}{\sqrt{2}}}{\frac{1}{u}\,du} \textrm{ after making the substitution } u = \cos{(x)} \\ &= -\frac{1}{2}\left[ e^{-2 \left( \frac{\pi}{4} \right) } - e^{2 \left( 0 \right) } \right] - 2 \left[ \ln{|u|} \right] _1 ^{\frac{1}{\sqrt{2}}} \\ &= -\frac{1}{2}\left( e^{-\frac{\pi}{2}} - 1 \right) - 2\left[ \ln{\left( \frac{1}{\sqrt{2}} \right)} - \ln{(0)} \right] \\ &= -\frac{1}{2} \left( e^{-\frac{\pi}{2}} - 1 \right) - 2\ln{ \left( 2^{-\frac{1}{2}} \right) } \\ &= -\frac{1}{2} \left( e^{-\frac{\pi}{2}} - 1 \right) + \ln{(2)} \end{align*}


    5.
    \displaystyle \begin{align*} \int_0^{\frac{\pi}{4}}{\tan^3{(x)}\,dx} &= \int_0^{\frac{\pi}{4}}{ \tan{(x)}\tan^2{(x)} \, dx } \\ &= \int_0^{\frac{\pi}{4}}{\tan{(x)}\left[ \sec^2{(x)} - 1 \right] dx } \\ &= \int_0^{\frac{\pi}{4}}{ \tan{(x)}\sec^2{(x)} - \tan{(x)}\,dx } \\ &= \int_0^{\frac{\pi}{4}}{ \tan{(x)}\sec^2{(x)}\,dx } + \int_0^{\frac{\pi}{4}}{\frac{-\sin{(x)}}{\cos{(x)}}\,dx} \\ &= \int_0^1{u\,du} + \int_1^{\frac{1}{\sqrt{2}}}{\frac{1}{v}\,dv} \textrm{ after making the substitutions } u = \tan{(x)} \implies du = \sec^2{(x)}\,dx \textrm{ and } v = \cos{(x)} \implies dv = -\sin{(x)}\,dx \\ &= \left[ \frac{u^2}{2} \right]_0^1 + \left[ \ln{|v|} \right]_1^{\frac{1}{\sqrt{2}}} \\ &= \left( \frac{1^2}{2} - \frac{0^2}{2} \right) + \left[ \ln{ \left| \frac{1}{\sqrt{2}} \right|} - \ln{|1|} \right] \\ &= \frac{1}{2} - \frac{1}{2}\ln{(2)} \end{align*}
    Thanks from alexander9408
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