Question

evaluate the integral.
int 25\tan^{5}xsec^{4}x dx
int 25\tan^{5}xsec^{4}x dx=square

Explanation:

Step1: Rewrite $\sec^{4}x$

We know that $\sec^{4}x=\sec^{2}x\cdot\sec^{2}x$. And $\sec^{2}x = 1+\tan^{2}x$. So the integral $\int25\tan^{5}x\sec^{4}x dx=25\int\tan^{5}x(1 + \tan^{2}x)\sec^{2}x dx$.

Step2: Use substitution

Let $u = \tan x$, then $du=\sec^{2}x dx$. The integral becomes $25\int u^{5}(1 + u^{2})du=25\int(u^{5}+u^{7})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $25(\int u^{5}du+\int u^{7}du)=25(\frac{u^{6}}{6}+\frac{u^{8}}{8})+C$.

Step4: Substitute back $u=\tan x$

We get $25(\frac{\tan^{6}x}{6}+\frac{\tan^{8}x}{8})+C=\frac{25\tan^{6}x}{6}+\frac{25\tan^{8}x}{8}+C$.

Answer:

$\frac{25\tan^{6}x}{6}+\frac{25\tan^{8}x}{8}+C$