All hell to integration

[BLUE_CHIP]

OK. I've had a little break from my studdies and need some help with this...

[latex]I_n(x)=\int\limits_0^x \tan^n{{\theta}}{{d\theta}},n\leq{0},{{x}}<\frac{\pi}{2}[\latex]

By writing [latex]\tan{\theta}[\latex] as [latex]\tan^{n-2}{\theta}\tan^2{\theta}[\latex], or otherwise, show that

[latex]I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x<\frac{\pi}{2}[\latex]

Hence evaluate [latex]\int\limits_{0}^{\frac{\pi}{3}}\tan^4{\theta}d\the ta[\latex], leaving your answers in terms of [latex]\pi[\latex]

Thanks (Goddam further maths)

AHHHH some one edit my post and get this bloody tex to work!!! pls

[Ambitwistor]

I_n(x)=\int\limits_0^x \tan^n \theta\,d\theta, n\leq{0}, x<\frac{\pi}{2}

By writing \tan{\theta} as \tan^{n-2}{\theta}\tan^2{\theta}, or otherwise, show that

I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x<\frac{\pi}{2}

Hence evaluate \int_0^{\pi/3}\tan^4{\theta}\,d\theta, leaving your answers in terms of \pi

[Hurkyl]

replace [ latex ] with [ tex ] (no spaces)