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Section 6.7

By cssp

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) f'(x) = 3\cosh(3x)

(c) f'(t) = 6t\sinh(t)\sinh(2t) + 3t\cosh(t)\cosh(2t) + 3\sinh(t)\cosh(t)

(e) y'(t) = 40t^2\cosh(4t)\sinh(4t) + 10t\cosh^2(4t)

2. (a) \displaystyle{\int \sinh(3x)dx = \frac{1}{3}\cosh(3x) + c}

(c) \displaystyle{\int \sinh(z)\cosh(z)dz = \frac{1}{2}\sinh^2(z) + c}

(e) \displaystyle{\int e^{-2t}\cosh(2t)dt = \frac{1}{2}t - \frac{1}{8}e^{-4t} + c}

(g) \displaystyle{\int 5t^2\cosh(2t)dt = \frac{5}{2}t^2\sinh(2t) - \frac{5}{2}t\cosh(2t) + \frac{5}{4}\sinh(2t) + c}

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

(c) \displaystyle{\int \frac{3}{\sqrt{9+3t^2}} \ dt = \sqrt{3}\sinh^{-1}\left(\frac{t}{\sqrt{3}}\right) + c}

(d) For x < -1 , \displaystyle{\int \frac{1}{\sqrt{x^2-1}} \ dx = -\cosh^{-1}|x| + c}.

6. (a) \displaystyle{f'(x) = 12x\mathrm{sech}^2(4x) + 3\tanh(4x)}

(c) \displaystyle{h'(\theta) = 8\tanh(\theta)\mathrm{sech}^3(\theta) - 4\tanh^3(\theta)\mathrm{sech}(\theta)}

7. (a) \displaystyle{\int \tanh(x)dx = \log(\cosh(x)) + c}

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

This entry was posted on 10 August 10 2007 at 12:48 pm and is filed under Section 6.7. 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.

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