Please don't mark my question as a duplicate of Find the reduction formula for the following integral.
This question, was asked by a user and I was trying to answer this, I couldn't complete my work and I don't know how to complete it. So I decided to ask my question requesting for help to see where I went wrong and how do I continue (since My working doesn't seem to end).
My working:
$$J_n=\tan^{2n} x \sec^3 x dx$$ $$J_n=\frac{1}{2n+1} \int \left(\frac{d}{dx} \tan^ {2n+1} x\right)\sec x dx$$
$$J_n=\frac{1}{2n+1} \left[\tan^{2n+1} x \sec x \right]-\frac{1}{2n+1} \int \tan^{2n+1} x \sec x \tan x dx$$
$$J_n=\frac{1}{2n+1} \left[\tan^{2n+1} x \sec x \right]-\frac{1}{2n+1} \int \tan^{2n} x \sec x (\sec^2 x -1) dx$$
$$J_n=\frac{1}{2n+1} \left[\tan^{2n+1} x \sec x \right]-\frac{1}{2n+1} \int \tan^{2n} x \sec^3 x dx+\frac{1}{2n} \int \tan^{2n} x \sec x dx$$
$$J_n=\frac{1}{2n+1} \left[\tan^{2n+1} x \sec x \right]-\frac{1}{2n+1} J_n+\frac{1}{2n} \int \tan^{2n-2} x \sec x (\sec^2 x -1) dx$$
$$\frac{2n}{2n+1} J_n=\frac{1}{2n+1} [\tan^{2n+1} x \sec x ]+\frac{1}{2n} \int \tan^{2n-2}x\sec^3 x dx -\frac{1}{2n} \int \tan^{2n-2} x \sec x dx$$
$$\frac{2n}{2n+1} J_n=\frac{1}{2n+1} [\tan^{2n+1} x \sec x ]+\frac{1}{2n} J_{n-1} -\frac{1}{2n} \int \tan^{2n-2} x \sec x dx$$
How do I find $\int \tan^{2n-2} x \sec x dx$ To complete this ?
I don't see an end to my working
Please help me to complete
Everything up to the fourth line seems correct. I disagree with your fifth line: I think it should read $$J_n = \frac{\tan^{2n+1} x \sec x}{2n+1} - \frac{J_n}{2n+1} + \frac{1}{2n+1} \int \tan^{2n} x \sec x \, dx.$$ In fact, it is easier to simply multiply everything by $2n+1$ after the second step to obtain $$\tag{*}2(n+1)J_n = \tan^{2n+1} x \sec x + \int \tan^{2n} x \sec x \, dx.$$ Now as for evaluating this last integral, recall that you wrote in the third line the equivalent of $$(2n+1)J_n = \tan^{2n+1} x \sec x - \int \tan^{2(n+1)} x \sec x \, dx,$$ so shifting the index of $n$ down by $1$ gives $$\tag{**}\int \tan^{2n} x \sec x \, dx = \tan^{2n-1} x \sec x - (2n-1) J_{n-1}.$$ This gives us the desired integral in terms of $J_{n-1}$, and a path toward a reduction formula. Consequently, after substituting $(**)$ into $(*)$, we obtain $$2(n+1)J_n = \tan^{2n+1} x \sec x + \tan^{2n-1} x\sec x - (2n-1)J_{n-1},$$ or equivalently, $$J_n = \frac{\tan^{2n-1} x \sec^3 x}{2(n+1)} - \frac{2n-1}{2(n+1)}J_{n-1}.$$