# All hell to integration

1. Dec 1, 2003

### 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] By writing [itex]\tan{\theta}[\latex] as [itex]\tan^{n-2}{\theta}\tan^2{\theta}[\latex], or otherwise, show that [itex]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 [itex]\int\limits_{0}^{\frac{\pi}{3}}\tan^4{\theta}d\theta[\latex], leaving your answers in terms of [itex]\pi[\latex] Thanks (Goddam further maths) AHHHH some one edit my post and get this bloody tex to work!!! pls Last edited by a moderator: Dec 1, 2003 2. Dec 1, 2003 ### Ambitwistor $$I_n(x)=\int\limits_0^x \tan^n \theta\,d\theta, n\leq{0}, x<\frac{\pi}{2}$$ By writing [itex]\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 $\pi$

Last edited: Dec 1, 2003
3. Dec 1, 2003

### Hurkyl

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