Question

evaluate the integral.
int sec^{2}5x\tan^{2}5x dx
int sec^{2}5x\tan^{2}5x dx=square

Explanation:

Step1: Use substitution

Let $u = \tan(5x)$. Then $du=5\sec^{2}(5x)dx$, and $\sec^{2}(5x)dx=\frac{1}{5}du$. The integral $\int\sec^{2}(5x)\tan^{2}(5x)dx$ becomes $\frac{1}{5}\int u^{2}du$.

Step2: Integrate $u^{2}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), for $n = 2$, we have $\frac{1}{5}\int u^{2}du=\frac{1}{5}\times\frac{u^{3}}{3}+C$.

Step3: Substitute back $u$

Substituting $u=\tan(5x)$ back into the result, we get $\frac{\tan^{3}(5x)}{15}+C$.

Answer:

$\frac{\tan^{3}(5x)}{15}+C$