Prove that $\int{\sec^{n}(\theta)}=\frac{\tan(\theta)\sec^{n-2}(\theta)}{n-1}-\frac{n-2}{n-1}\int{\sec^{n-2}(\theta)d\theta}$

Question

One of the problems in my homework set ask me to prove the following identity: $$\int{\sec^{n}(\theta)}d\theta=\frac{\tan(\theta)\sec^{n-2}(\theta)}{n-1}-\frac{n-2}{n-1}\int{\sec^{n-2}(\theta)d\theta}$$ Can someone give me a hint how to start? Thanks!

Answer 1

Using integration by parts, $$\int\sec^nxdx=\int\sec^{n-2}x\cdot \sec^2xdx$$ $$=\sec^{n-2}x\int\sec^2xdx-\int\left(\frac{d(\sec^{n-2}x)}{dx}\cdot \sec^2xdx\right)dx $$

$$=\sec^{n-2}x\tan x-\int\left((n-2)\sec^{n-3}x(\sec x\tan x)\cdot \tan x\right)dx $$

$$=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x(\sec^2x-1)dx $$

$$=\sec^{n-2}x\tan x-(n-2)\int\sec^nxdx+(n-2)\int\sec^{n-2}xdx+C $$ where $C$ is an arbitrary constant for indefinite integration

$$\implies (1+n-2)\sec^nxdx=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}xdx+C$$

Answer 2

Hint:

Use the fact that

$$\tan^2 (\theta)+1=\sec^2(\theta)$$

Now , $\sec^n \theta=\sec^{n-2} \theta \cdot \sec^2 \theta=\sec^{n-2}\theta \cdot (\tan^2 \theta+1)$

Answer 3

Hint:

$$\int{\sec^{n}(\theta)}d\theta= \int{\sec^{n-2}(\theta)} \sec^2(\theta) d\theta$$

Integrate by parts.

Answer 4

Hint:

remember that $d(\tan \theta) = \frac{d\theta}{\cos^2\theta} = \sec^2 \theta d\theta$, integrate by parts, then collect the like terms with $\int \frac{d\theta}{\cos^n \theta}$