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Math Help - Integral

  1. #1
    Junior Member
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    Integral

    Hi,

    Can you help me to calculate the following integral: \int_0^{\frac{\pi}{4}} tan^3(t)+tan(t) dt= \int_0^{\frac{\pi}{4}}tan(t)(1+tan^2(t))dt \ = \ \int_0^{\frac{\pi}{4}}tan(t)tan'(t) dt=???
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by lehder View Post
    Hi,

    Can you help me to calculate the following integral: \int_0^{\frac{\pi}{4}} tan^3(t)+tan(t) dt= \int_0^{\frac{\pi}{4}}tan(t)(1+tan^2(t))dt \ = \ \int_0^{\frac{\pi}{4}}tan(t)tan'(t) dt=???
    Derivative of (\tan(x))^2 is...
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  3. #3
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    Hello, lehder!

    You're almost there . . .


    \int_0^{\frac{\pi}{4}} \left(\tan^3t+\tan t\right)\,dt

    We have: . \int \tan x\left(\tan^2t + 1\right)\,dt \;=\;\int\tan x\,\sec^2\!x\,dx


    Let: u \,=\,\tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx

    Substitute: . \int u\,du \;=\;\tfrac{1}{2}u^2+C

    Back-substitute: . \tfrac{1}{2}\tan^2\!x + C


    Evaluate: . \tfrac{1}{2}\tan^2\!x\,\bigg]^{\frac{\pi}{4}}_0 \;=\;\tfrac{1}{2}\tan^2\!\tfrac{\pi}{4} - \tfrac{1}{2}\tan^2\!0 \;=\;\tfrac{1}{2}(1^2) - \tfrac{1}{2}(0^2) \;=\;\frac{1}{2}

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