- #1
BLUE_CHIP
OK. I've had a little break from my studdies and need some help with this...
I_n(x)=\int\limits_0^x \tan^n{{\theta}}{{d\theta}},n\leq{0},{{x}}<\frac{\pi}{2}[\latex]<br /> <br /> By writing \tan{\theta}[\latex] as \tan^{n-2}{\theta}\tan^2{\theta}[\latex], or otherwise, show that&lt;br /&gt; &lt;br /&gt; I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x&amp;amp;lt;\frac{\pi}{2}[\latex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Hence evaluate \int\limits_{0}^{\frac{\pi}{3}}\tan^4{\theta}d\theta[\latex], leaving your answers in terms of \pi[\latex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Thanks (Goddam further maths)&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; AHHHH some one edit my post and get this bloody tex to work! pls
I_n(x)=\int\limits_0^x \tan^n{{\theta}}{{d\theta}},n\leq{0},{{x}}<\frac{\pi}{2}[\latex]<br /> <br /> By writing \tan{\theta}[\latex] as \tan^{n-2}{\theta}\tan^2{\theta}[\latex], or otherwise, show that&lt;br /&gt; &lt;br /&gt; I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x&amp;amp;lt;\frac{\pi}{2}[\latex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Hence evaluate \int\limits_{0}^{\frac{\pi}{3}}\tan^4{\theta}d\theta[\latex], leaving your answers in terms of \pi[\latex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Thanks (Goddam further maths)&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; AHHHH some one edit my post and get this bloody tex to work! pls
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