I took the following route:

$$\int{\tan^6{x} \sec^4{x} \, \mathrm dx}$$ $$= \int{\tan^6{x} (\sec^2{x})^2 \, \mathrm dx}$$ $$= \int{\tan^6{x} (1 + \tan^2{x})^2 \, \mathrm dx}$$ $$= \int{\tan^6{x} (1 + 2\tan^2{x} + \tan^4{x})^2 \, \mathrm dx}$$ $$= \int{\tan^6{x} \, \mathrm dx} + 2\int{\tan^8{x} \, \mathrm dx} + \int{\tan^{10}{x} \, \mathrm dx} \tag{1}\label{1}$$

Focusing on just the first integral, I broke it up:

$$\int{\tan^6{x} \, \mathrm dx} \tag{2}\label{2}$$ $$= \int{\tan^4{x} \tan^2{x} \, \mathrm dx}$$ $$= \int{\tan^4{x} (\sec^2{x}-1) \, \mathrm dx}$$ $$ = \int{\tan^4{x} \sec^2{x} \, \mathrm dx} - \int{\tan^4{x}} \, \mathrm dx \tag{3}\label{3}$$

The first integral in $\eqref{3}$ can be evaluated using substitution, and the second can be evaluated using the same process that I used originally for $\eqref{2}$. The remaining two integrals in $\eqref{1}$ can be evaluated in the same manner.

After following this tedious, recursive process, I finally ended up with

$$\frac{1}{9} \tan^9{x} + \frac{1}{7} \tan^7{x} + C$$

Which is the correct answer. However, I would love to see a simpler solution. Thanks!


You have:$$\int\tan^6x\sec^4x\,\mathrm dx=\int\tan^6x(1+\tan^2x)\sec^2x\,\mathrm dx.$$Now, doing $\tan x=y$ and $\sec^2x\,\mathrm dx=\mathrm dy$, this becomes$$\int y^6+y^8\,\mathrm dy$$which is equal to$$\frac17y^7+\frac19y^9=\frac17\tan^7x+\frac19\tan^9x.$$


Better to pull out a $\sec^2 x$ and convert everything else to $\tan x$:

$$=\int \tan^6 x (1+\tan^2 x) \sec^2 x \; dx = \int u^6(1+u^2) \; du $$ $$= \frac{u^7}{7}+\frac{u^9}{9} + C = \frac{\tan^7 x}{7} + \frac{\tan^9 x}{9} +C.$$


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