evaluate the integral.\n int 6\tan^{5}xsec^{2}xmathrm{d}x\n int 6\tan^{5}xsec^{2}xmathrm{d}x=square

Explanation:

Step1: Use substitution

Let $u = \tan x$, then $du=\sec^{2}x dx$. The integral $\int 6\tan^{5}x\sec^{2}x dx$ becomes $\int 6u^{5}du$.

Step2: Integrate with power - rule

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int 6u^{5}du$, we have $6\times\frac{u^{6}}{6}+C$.

Step3: Simplify and substitute back

$6\times\frac{u^{6}}{6}+C=u^{6}+C$. Substituting $u = \tan x$ back, we get $\tan^{6}x + C$.

Answer:

$\tan^{6}x + C$