How Can We Simplify Integration Using Trigonometric Identities?

[BLUE_CHIP]

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

[itex]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

 

[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]

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