How Does the Reduction Formula Simplify Integration of Trigonometric Functions?

[John O' Meara]

Show that $$tan^n(x) = tan^{n-2}(x)(sec^2(x)-1) \\$$. Hence if $$I_n = \int_0^{\frac{\pi}{4}}tan^n(x)dx \\$$. Prove that $$I_n = \frac{1}{n-1} - I_{n-2} \\$$, and evaluate [tex]I_5{/tex]<br /> My effort:<br /> $$I_n = \int_0^{n-2}tan^{n-2}x(sec^2x-1)dx \\$$ $$du = (n-2)tan^{n-3}x sec^2x dx \\$$, and $$v = \int(sec^2x - 1) dx = tanx -x \\$$. Therefore: <br /> $$I_n = (tan^{n-2}x(tanx - x)) - (n-2)\int_0^{\frac{\pi}{4}}tan^{n-3}xsec^2x(tanx - x)dx \\$$.<br /> Which implies, $$I_n = (tan^{n-2}x(tanx - x) - (n-2)\int_0^{\frac{\pi}{4}}tan^{n-2}sec^2x dx - (n-2)\int tan^{n-3}x(sec^2x -1) dx \\$$. Therefore that implies :$$I_n = (tan^{n-1}x - tan^{n-2}(x) x) -(n-1)\int tan^{n-2}x(sec^2x -1)dx + (n-1)\int xtan^{n-3}x(sec^2x-1)dx \\$$<br /> <br /> The end result I get is : $$I_n = 1 - \frac{\pi}{4} - (n-1)I_n + (n-1)\int_0^{\frac{\pi}{4}}tan^{n-2}(x) x dx$$. I must have gone wrong some where to get this result. Thanks for the help.[/tex]

 

[AlephZero]

The first identity is true because $$\sec^2 x = 1 + \tan^2 x$$

For the integral, use that identity:

$$I_n = \int_0^{\frac \pi 4} tan^{n-2}x(sec^2x-1) dx$$
$$I_n = \int_0^{\frac \pi 4} tan^{n-2}xsec^2xdx - I_{n-2}$$

Evaluating the integral is straightforward - let u = tan x.