Question

evaluate $int \tan^{4}xsec^{6}x dx$

Explanation:

Step1: Rewrite using identity

We know that $\sec^{2}x = 1+\tan^{2}x$. So, $\int\tan^{4}x\sec^{6}x dx=\int\tan^{4}x\sec^{4}x\sec^{2}x dx=\int\tan^{4}x(1 + \tan^{2}x)^{2}\sec^{2}x dx$.

Step2: Use substitution

Let $u=\tan x$, then $du=\sec^{2}x dx$. The integral becomes $\int u^{4}(1 + u^{2})^{2}du$.

Step3: Expand the integrand

Expand $(1 + u^{2})^{2}=1 + 2u^{2}+u^{4}$. So the integral is $\int u^{4}(1 + 2u^{2}+u^{4})du=\int(u^{4}+2u^{6}+u^{8})du$.

Step4: Integrate term - by - term

Using the power rule $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int(u^{4}+2u^{6}+u^{8})du=\frac{u^{5}}{5}+\frac{2u^{7}}{7}+\frac{u^{9}}{9}+C$.

Step5: Substitute back

Substitute $u = \tan x$ back into the result. We get $\frac{\tan^{5}x}{5}+\frac{2\tan^{7}x}{7}+\frac{\tan^{9}x}{9}+C$.

Answer:

$\frac{\tan^{5}x}{5}+\frac{2\tan^{7}x}{7}+\frac{\tan^{9}x}{9}+C$