evaluate the integral.\n intsec^{2}2x\tan^{3}2x dx\n intsec^{2}2x\tan^{3}2x dx=square

Explanation:

Step1: Use substitution

Let $u = \tan2x$, then $du=2\sec^{2}2x dx$, and $\sec^{2}2x dx=\frac{1}{2}du$.

Step2: Rewrite the integral

The integral $\int\sec^{2}2x\tan^{3}2x dx$ becomes $\frac{1}{2}\int u^{3}du$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $\frac{1}{2}\int u^{3}du=\frac{1}{2}\times\frac{u^{4}}{4}+C=\frac{u^{4}}{8}+C$.

Step4: Substitute back

Substitute $u = \tan2x$ back into the result, we get $\frac{\tan^{4}2x}{8}+C$.

Answer:

$\frac{\tan^{4}2x}{8}+C$