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Thread: Trigonometric integration

  1. #1
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    Question Trigonometric integration

    Here's the question :

    Evaluate the foll.intgral

    [Int] 2 x^2 . cos (1/2 x^3 ) dx

    I tried to do it by u-substitution and got this answer :


    [4/3 x^3 . 4 sin (1/2 x^3 ) . 2 x^-2] + c

    This does not look right.
    Can someone pls show me how to do it?
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  2. #2
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    Hi

    Quote Originally Posted by vdemas View Post
    Here's the question :

    Evaluate the foll.intgral

    [Int] 2 x^2 . cos (1/2 x^3 ) dx

    I tried to do it by u-substitution and got this answer :


    [4/3 x^3 . 4 sin (1/2 x^3 ) . 2 x^-2] + c

    This does not look right.
    Can someone pls show me how to do it?
    $\displaystyle I=\int 2x^2 \cos \left(\frac 12 \cdot x^3 \right) dx$

    That's correct, make a substitution : $\displaystyle u=\frac 12 x^3$

    --> $\displaystyle du=\frac 32 x^2 dx \implies dx=\frac 2{3x^2} du$

    $\displaystyle \implies I=\int 2{\color{red}x^2} \cdot \cos u \cdot \frac 2{3{\color{red}x^2}} du=\frac 43 \int \cos u du$

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  3. #3
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    well if you dont want to use substitution you can recognize from the differentiation of sine functions:

    $\displaystyle
    \frac {d}{d\theta}sin(\theta) = cos(\theta)
    $

    $\displaystyle
    \frac {d}{d\theta}sin(\theta ^2) = 2\theta \cdot cos(\theta ^2)
    $

    now lets look at your case:
    $\displaystyle
    I=\int 2x^2 \cos \left(\frac 12 \cdot x^3 \right) dx
    $

    looking at the trigonometry what do you differentiate to obtain:
    $\displaystyle
    \cos \left(\frac 12 \cdot x^3 \right)
    $

    well you can see that $\displaystyle \sin \left(\frac 12 \cdot x^3 \right) $ will give you the required cos function:

    $\displaystyle
    \frac{d}{dx} \sin \left(\frac 12 \cdot x^3 \right) = \frac 32 x^2\cos \left(\frac 12 \cdot x^3 \right)
    $

    From here we can see that differentiating $\displaystyle \sin \left(\frac 12 \cdot x^3 \right)$ will give us almost the required answer.
    Multiplying both side with 4/3 gives:

    $\displaystyle
    \frac 43 \cdot \frac{d}{dx} \sin \left(\frac 12 \cdot x^3 \right) = \frac43 \cdot \frac 32 x^2\cos \left(\frac 12 \cdot x^3 \right)
    $
    $\displaystyle
    \frac{d}{dx} \frac 43 \sin \left(\frac 12 \cdot x^3 \right) = 2 x^2\cos \left(\frac 12 \cdot x^3 \right)
    $

    Now we just integrate both sides to obtain the required question:
    $\displaystyle
    \int \frac{d}{dx} \frac 43 \sin \left(\frac 12 \cdot x^3 \right) = \int 2 x^2\cos \left(\frac 12 \cdot x^3 \right)
    $
    $\displaystyle
    \int 2 x^2\cos \left(\frac 12 \cdot x^3 \right)= \frac 43 \sin \left(\frac 12 \cdot x^3 \right)
    $
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