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Section 6.6

Please post comments if you find any errors, or if you wish to contribute answers for additional problems. You may find information on using \LaTeX here.

1. (a) \displaystyle{\int \frac{3}{\sqrt{16-x^2}} \ dx = 3\sin^{-1}\left(\frac{x}{4}\right) + c}

(c) \displaystyle{\int \sqrt{5-z^2} \ dz = \frac{z}{2}\sqrt{5-z^2} + \frac{5}{2}\sin^{-1}\left(\frac{z}{\sqrt{5}}\right) + c}

(e) \displaystyle{\int \frac{1}{\sqrt{4+x^2}} \ dx = \log\left|\sqrt{4+x^2} + x\right| + c}

2. (a) \displaystyle{\int z\sqrt{1-z^2} \ dz = -\frac{1}{3}(1 - z^2)^{\frac{3}{2}} + c}

(c) \displaystyle{\int \frac{1}{\sqrt{x^2 - 4}} \ dx = \log\left|x + \sqrt{x^2-4}\right| + c}

(e) \displaystyle{\int \frac{4}{\sqrt{3-2x^2}} \ dx = 2\sqrt{2}\sin^{-1}\left(\sqrt{\frac{2}{3}}x\right) + c}

(g) \displaystyle{\int \sqrt{4-t^2} \ dt = \frac{t}{2}\sqrt{4-t^2} + 2\sin^{-1}\left(\frac{t}{2}\right)}

3. (a) \displaystyle{\int_0^3 \sqrt{9-t^2} \ dt = \frac{9\pi}{4}}

(c) \displaystyle{\int_0^1 \frac{1}{(1+x^2)^{\frac{3}{2}}} \ dx = \frac{1}{\sqrt{2}}}

(e) \displaystyle{\int_{5\sqrt{2}}^{10} \frac{1}{\sqrt{x^2-25}} \ dx = \log(2+\sqrt{3}) - \log(1 + \sqrt{2})}

5. \displaystyle{\int \frac{1}{\sqrt{1-x^2}} \ dx = -\cos^{-1}(x) + c}

This entry was posted on 10 August 10 2007 at 8:31 am and is filed under Section 6.6. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Section 6.6”

  1. Juan Manuel Barreneche Says:
    26 December 26 2007 at 2:06 pm

    The answer for 1.e it’s wrong: me and wolfram’s integrator (http://integrals.wolfram.com/index.jsp) agree that the solution is

    log|sqrt(x² + 4)/2 + x/2| + C

    thank you,

    JM

    PD: The book it’s great.

  2. Juan Manuel Barreneche Says:
    26 December 26 2007 at 2:15 pm

    sorry, i was wrong,

    log|sqrt(x² + 4)/2 + x/2| + C
    = log(|sqrt(x² + 4) + x|/2) + C
    = log|sqrt(x² + 4) + x| - log(2) + C
    = log|sqrt(x² + 4) + x| + C1
    were C1 = C - log(2)

    JM

  3. cssp Says:
    26 December 26 2007 at 2:26 pm

    That’s right. Integrals can be tricky that way.

    Dan Sloughter

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