evaluate the integral.\n int 3sec^{4}x\tan^{7}x dx\n int 3sec^{4}x\tan^{7}x dx=square

Explanation:

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

Recall $\sec^{2}x = 1+\tan^{2}x$, so $\sec^{4}x=\sec^{2}x\cdot\sec^{2}x=(1 + \tan^{2}x)\sec^{2}x$. Then the integral $\int3\sec^{4}x\tan^{7}x dx=3\int(1 + \tan^{2}x)\sec^{2}x\tan^{7}x dx$.

Step2: Use substitution

Let $u = \tan x$, then $du=\sec^{2}x dx$. The integral becomes $3\int(1 + u^{2})u^{7}du=3\int(u^{7}+u^{9})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $3\int(u^{7}+u^{9})du=3(\frac{u^{8}}{8}+\frac{u^{10}}{10})+C$.

Step4: Substitute back $u=\tan x$

The result is $3(\frac{\tan^{8}x}{8}+\frac{\tan^{10}x}{10})+C=\frac{3\tan^{8}x}{8}+\frac{3\tan^{10}x}{10}+C$.

Answer:

$\frac{3\tan^{8}x}{8}+\frac{3\tan^{10}x}{10}+C$