Question

evaluate $int sec^{3}x\tan^{3}x dx$

Explanation:

Step1: Rewrite $\tan^{3}x$

We know that $\tan^{3}x=\tan^{2}x\tan x=(\sec^{2}x - 1)\tan x$. So the integral becomes $\int\sec^{3}x(\sec^{2}x - 1)\tan xdx=\int(\sec^{5}x\tan x-\sec^{3}x\tan x)dx$.

Step2: Use substitution

Let $u = \sec x$, then $du=\sec x\tan xdx$.
For $\int\sec^{5}x\tan xdx$, substituting $u = \sec x$ gives $\int u^{4}du$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$, we have $\int u^{4}du=\frac{u^{5}}{5}+C_1=\frac{\sec^{5}x}{5}+C_1$.
For $\int\sec^{3}x\tan xdx$, substituting $u=\sec x$ gives $\int u^{2}du$. By the power - rule, $\int u^{2}du=\frac{u^{3}}{3}+C_2=\frac{\sec^{3}x}{3}+C_2$.

Step3: Combine the results

$\int(\sec^{5}x\tan x-\sec^{3}x\tan x)dx=\frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C$ (where $C = C_1 - C_2$).

Answer:

$\frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C$