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Math Help - integration

  1. #1
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    integration

    integrate the following using trigonometric identities and/or substitutions:

    x^2 + 1 / sqrt (9 - x^2) dx

    x^3 / sqrt (x^2 + 4) dx

    (tan x)^5 * (sec x)^6 dx

    (cos x)^5 / sqrt (sin x) dx


    with the first two i started out confidently but ended up wasting half an hour trying o figure them out .... and the last two are just impossible ...
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by razorfever View Post
    (tan x)^5 * (sec x)^6 dx
    Note that \int\tan^5x\sec^6x\,dx=\int\tan^5x\left(\sec^2x\ri  ght)^2\sec^2x\,dx=\int\tan^5x\left(\tan^2x+1\right  )^2\sec^2x\,dx =\int\left(\tan^9x+2\tan^7x+\tan^5x\right)\sec^2x\  ,dx

    Now apply the substitution z=\tan x

    (cos x)^5 / sqrt (sin x) dx
    Note that \int\frac{\cos^5x\,dx}{\sqrt{\sin x}}=\int\frac{\left(\cos^2x\right)^2\cos x\,dx}{\sqrt{\sin x}}=\int\frac{\left(1-\sin^2x\right)^2\cos x\,dx}{\sqrt{\sin x}}

    Not apply the substitution z=\sin x

    Can you try these two now?
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  3. #3
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    Hello, razorfever!

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

    Let: x \:=\:3\sin\theta \quad\Rightarrow\quad dx \:=\:3\cos\theta\,d\theta \quad\Rightarrow\quad \sqrt{9=x^2} \:=\:3\cos\theta

    Substitute: . \int \frac{9\sin^2\!\theta +1}{3\cos\theta}(3\cos\theta\,d\theta) \;=\;\int\left(9\sin^2\!\theta + 1\right)\,d\theta

    Can you finish it?




    \int \frac{x^3}{\sqrt{x^2 + 4}}\,dx

    Let: x \:=\: 2\tan\theta \quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta \quad\Rightarrow\quad \sqrt{x^2+4} \:=\:2\sec\theta

    Substitute: . \int\frac{8\tan^3\!\theta}{2\sec\theta}\,\left(2\s  ec^2\!\theta\,d\theta\right) \;=\;8\int\sec\theta\tan^3\!\theta\,d\theta

    . . = \;8\int\tan^2\!\theta(\sec\theta\tan\theta\,d\thet  a)  \;=\;8\int(\sec^2\!\theta - 1)(\sec\theta\tan\theta\,d\theta)


    Now let: . u \:=\:\sec\theta

    Got it?

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  4. #4
    o_O
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    For the first one, use trig sub: x = 3\sin \theta \ \Rightarrow \ dx = 3\cos \theta

    _____________________

    For the second one, use the sub: u = x^2 + 4 \ \Rightarrow \frac{du}{2} = x \ dx

    and note that you can rearrange your sub to get: x^2 = u - 4

    This will give you: \int \frac{x^2 \cdot x \ dx}{\sqrt{x^2 + 4}} = \frac{1}{2}\int \frac{u-4}{u^{\frac{1}{2}}} \ du = \frac{1}{2}\int (u^\frac{1}{2} - 4u^{-\frac{1}{2}} ) du
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  5. #5
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    thanks
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