Question

evaluate the integral.
int 8\tan xsec^{7}x dx
int 8\tan xsec^{7}x dx=square

Explanation:

Step1: Use substitution method

Let $u = \sec x$, then $du=\sec x\tan xdx$. The integral $\int 8\tan x\sec^{7}x dx$ can be rewritten as $8\int\sec^{6}x(\sec x\tan xdx)$.

Step2: Substitute $u$

Since $u = \sec x$, the integral becomes $8\int u^{6}du$.

Step3: Integrate $u^{6}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $8\times\frac{u^{7}}{7}+C=\frac{8}{7}u^{7}+C$.

Step4: Substitute back $u$

Substituting $u=\sec x$ back, we get $\frac{8}{7}\sec^{7}x + C$.

Answer:

$\frac{8}{7}\sec^{7}x + C$