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Math Help - Integration by substitution

  1. #1
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    Question Integration by substitution

    I'm stuck on several questions that are driving me round the bend. If you can help with any it would be greatly appreciated.

    a)Integral of: 1/sqrtx(2+sqrtx).dx = ln 16, between the values of 36 and 0 and where the substitution is x=u^2

    or

    b)Integral of: cos^2(t)sin(t).dt, between the values of pi/2 and pi/3 where u = cos(t)

    sorry if it is hard to understand but I am new to this and don't understand how to insert the proper maths lingo and symbols.

    Thanks Beard
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by Beard View Post
    a)Integral of: 1/sqrtx(2+sqrtx).dx = ln 16, between the values of 36 and 0 and where the substitution is x=u^2
    I=\int_0^{36}\frac{1}{\sqrt{x}\left(2+\sqrt{x}\rig  ht)}\,\mathrm{d}x

    Let x=u^2. We have  \frac{\mathrm{d}x}{\mathrm{d}u}=2u hence  \mathrm{d}x=2u\,\mathrm{d}u.

    I=\int_{\sqrt{0}}^{\sqrt{36}} \frac{1}{\sqrt{u^2}\left(2+\sqrt{u^2}\right)}\cdot 2u\,\mathrm{d}u

    Since u\geq0,\, \sqrt{u^2}=|u|=u so
    <br />
\begin{aligned}<br />
I&=2\int_0^6 \frac{1}{u\left(2+u\right)}\cdot u\,\mathrm{d}u\\<br />
&=2\int_0^6 \frac{1}{2+u}\,\mathrm{d}u\\<br />
\end{aligned}

    Can you take it from here ?
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  3. #3
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    Thumbs up

    Merci

    I can see where I was going wrong
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Beard View Post
    Merci
    De rien.

    Quote Originally Posted by Beard View Post
    b)Integral of: cos^2(t)sin(t).dt, between the values of pi/2 and pi/3 where u = cos(t)
    u=\cos t so \frac{\mathrm{d}u}{\mathrm{d}t}=-\sin t \implies  \mathrm{d}u=-\sin(t)\,\mathrm{d}t.

    \int_{\pi/2}^{\pi/3}\underbrace{\cos^2(t)}_{u^2}\underbrace{\sin(t)\  ,\mathrm{d}t}_{-\mathrm{d}u}=\int_{\cos(\pi/3)}^{\cos(\pi/2)}u^2\mathrm{d}u=\ldots
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  5. #5
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    Thumbs up

    Thanks again
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