MHB How Do You Use Reduction Formulae for tan^n x?

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The discussion focuses on the application of reduction formulae for integrating \( \tan^n x \). An example is provided using \( n=4 \), demonstrating that \( \int \tan^4 x \, dx \) can be expressed as \( \frac{1}{3} \tan^3 x - \int \tan^2 x \, dx \). The integration of \( \tan^2 x \) is further simplified using the identity \( \tan^2 x = \sec^2 x - 1 \). The final result of the integration is \( \frac{1}{3} \tan^3 x - \tan x + x + C \). This illustrates how reduction formulae can simplify the integration of higher powers of tangent functions.
shamieh
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Can someone show me how I would "use" the reduction formulae for $$tan^n x$$ ? I just want to see an example on when I would ever use it. A simple one will do.$$\frac{tan^{n-2}x}{n - 1} - \int tan^{n-2} x dx$$
 
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Hello, shamieh!

Your formula is incorrect.

\int\tan^n\!x\,dx \;=\;\frac{\tan^{n-1}\!x}{n - 1} - \int\tan^{n-2}\!x\,dxExample: .\int\tan^4\!x\,dx

Substitute n=4 into the formula:

\int\tan^4\!x\,dx \;=\;\tfrac{1}{3}\tan^3\!x - \int \tan^2\!x\,dx

. . . . . . . . . =\;\tfrac{1}{3}\tan^3\!x - \int(\sec^2\!x -1)\,dx

. . . . . . . . . =\;\tfrac{1}{3}\tan^3\!x - \int\sec^2\!x\,dx + \int dx

. . . . . . . . . =\;\tfrac{1}{3}\tan^3\!x - \tan x + x + C
 
Thank you sororaban! Also thank you for the PM the other day!
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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