Question

evaluate the integral.
int sec^{7}5x\tan5x dx
int sec^{7}5x\tan5x dx=square

Explanation:

Step1: Use substitution

Let $u = \sec5x$. Then $du=5\sec5x\tan5x dx$, and $\sec5x\tan5x dx=\frac{1}{5}du$. The integral $\int\sec^{7}5x\tan5x dx$ becomes $\frac{1}{5}\int u^{6}du$.

Step2: Apply power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $\frac{1}{5}\int u^{6}du$, we have $\frac{1}{5}\times\frac{u^{7}}{7}+C$.

Step3: Substitute back $u$

Substitute $u = \sec5x$ back into the result. We get $\frac{\sec^{7}5x}{35}+C$.

Answer:

$\frac{\sec^{7}5x}{35}+C$