Question

evaluate the integral.
int 3\tan xsec^{7}x dx
int 3\tan xsec^{7}x dx=square

Explanation:

Step1: Use substitution

Let \(u = \sec x\), then \(du=\sec x\tan xdx\). The integral \(\int3\tan x\sec^{7}x dx = 3\int\sec^{6}x\cdot\sec x\tan xdx\). After substitution, it becomes \(3\int u^{6}du\).

Step2: Integrate using power - rule

The power - rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(3\int u^{6}du\), we have \(3\times\frac{u^{7}}{7}+C=\frac{3}{7}u^{7}+C\).

Step3: Substitute back

Substitute \(u = \sec x\) back into the result. So the integral \(\int3\tan x\sec^{7}x dx=\frac{3}{7}\sec^{7}x + C\).

Answer:

\(\frac{3}{7}\sec^{7}x + C\)