Question

$\int \frac{(3n^4+3n^3)}{\cos x}dx$

Answer 1

Given the integral,

$\int \frac{3n^4+3n^3}{\cos x}dx=\int 3(n+1)n^3\sec xdx=3(n+1)n^3\int \sec xdx$

For $\int \sec xdx$,

$\int \sec xdx=\int \frac{\sec x(\tan x+\sec x)}{\tan x+\sec x}dx=\int \frac{\sec x\tan x+{\sec }^{2}x}{\tan x+\sec x}dx$

Let,

$u=\tan x+\sec x\Rightarrow \frac{du}{dx}={\sec }^{2}x+\sec x\tan x\Rightarrow du={\sec }^{2}x+\sec x\tan xdx$

So,

$\int \frac{\sec x\tan x+{\sec }^{2}x}{\tan x+\sec x}dx=\int \frac{1}{u}du=\ln \left(u\right)=\ln (\tan x+\sec x)\therefore 3(n+1)n^3\int \sec xdx=3(n+1)n^3\ln (\tan x+\sec x)$

Hence, $\int \frac{3n^4+3n^3}{\cos x}dx=3(n+1)n^3\ln \left(|\tan x+\sec x|\right)+C.$