Question

evaluate $intsec^{3}x\tan^{5}x dx$ +c

Explanation:

Step1: Rewrite $\tan^{5}x$

We know that $\tan^{5}x=\tan^{4}x\cdot\tan x = (\sec^{2}x - 1)^{2}\tan x$. So the integral becomes $\int\sec^{3}x(\sec^{2}x - 1)^{2}\tan xdx$.

Step2: Use substitution

Let $u = \sec x$, then $du=\sec x\tan xdx$. The integral $\int\sec^{3}x(\sec^{2}x - 1)^{2}\tan xdx=\int u^{2}(u^{2}-1)^{2}du$.

Step3: Expand the integrand

Expand $(u^{2}-1)^{2}=u^{4}-2u^{2}+1$. Then $u^{2}(u^{2}-1)^{2}=u^{2}(u^{4}-2u^{2}+1)=u^{6}-2u^{4}+u^{2}$.

Step4: Integrate term - by - term

$\int(u^{6}-2u^{4}+u^{2})du=\frac{u^{7}}{7}-2\times\frac{u^{5}}{5}+\frac{u^{3}}{3}+C$.

Step5: Substitute back $u = \sec x$

We get $\frac{\sec^{7}x}{7}-\frac{2\sec^{5}x}{5}+\frac{\sec^{3}x}{3}+C$.

Answer:

$\frac{\sec^{7}x}{7}-\frac{2\sec^{5}x}{5}+\frac{\sec^{3}x}{3}+C$