Aye gents, I've been working on a problem that doesn't have a solution at the back of the book and would be grateful if you'd compare my answer with yours.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

[tex]\int\frac{t^5}{\sqrt{t^2 + 2}}\ dt[/tex]

2. Relevant equations

[tex]t = \sqrt{2}\tan\theta[/tex]

[tex]dt = \sqrt{2}\sec^2\theta\ d\theta[/tex]

3. The attempt at a solution

[tex]\int\frac{(\sqrt{2}\tan\theta)^5}{\sqrt{(\sqrt{2}\tan\theta)^2 + 2}}\ \sqrt{2}\sec^2\theta\ d\theta[/tex]

[tex]\int\frac{8\tan^5\theta\sec^2\theta}{\sqrt{2\tan^2\theta + 2}\ d\theta}[/tex]

[tex]8\int\frac{\tan^5\theta\sec^2\theta}{\sqrt{2}\sec\theta}\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int\tan^5\theta\sec\theta\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int\frac{\sin^5\theta}{\cos^6\theta}\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int\frac{[(1 - \cos^2\theta)(1 - \cos^2\theta)]}{\cos^6\theta}\ \sin\theta\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int[\frac{1}{\cos^6\theta} - \frac{2}{\cos^4\theta} + \frac{1}{\cos^2\theta}]\ \sin\theta\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\ [\frac{-1}{7}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^7) + \frac{2}{5}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^5) - \frac{1}{3}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^3)]\ +\ C[/tex]

Thanks.

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# Homework Help: More Trig Subsitution

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