Integrating $\sec^{3}\theta$ .

Question

How do I integrate the trigonometric term ?

$$\sec^{3} \theta$$

My assumption

I tried to write it in cosine form and apply formula of $\cos^{3}\theta$ , but to no purpose. What should I do ?

Answer 1

You have\begin{align}\int\sec^3\theta\,\mathrm d\theta&=\int\sec^2\theta\sec\theta\,\mathrm d\theta\\&=\tan\theta\sec\theta-\int\tan\theta\sec'\theta\,\mathrm d\theta\\&=\tan\theta\sec\theta-\int\tan^2\theta\sec\theta\,\mathrm d\theta\\&=\tan\theta\sec\theta-\int\frac{\sin^2\theta}{\cos^3\theta}\,\mathrm d\theta\\&=\tan\theta\sec\theta-\int\frac{\cos\theta\sin^2\theta}{(1-\sin^2\theta)^2}\,\mathrm d\theta.\end{align}Now, you can compute this last primitive using the substitution $\sin\theta=x$ and $\cos\theta\,\mathrm d\theta=\mathrm dx$

Answer 2

use that $$\int\sec^m(x)dx=\frac{\sin(x)\sec^{m-1}(x)}{m-1}+\frac{m-2}{m-1}\int\sec^{m-2}(x)dx$$ and set $m=3$!

Answer 3

$$\int \sec^2 x \sec x \ \mathrm dx =\int (1+\tan^2 x) \sec^2 x \ \mathrm dx$$ Set $u=\tan x$ $$\int 1+u^2 \ \mathrm du= u+\frac{u^3}{3}+C = \tan x+\frac{\tan^3 x}{3}+C$$

Answer 4

Hint: $$ \begin{align} \int\sec^3(\theta)\,\mathrm{d}\theta &=\int\sec^4(\theta)\,\mathrm{d}\sin(\theta)\\ &=\int\frac1{\left(1-\sin^2(\theta)\right)^2}\,\mathrm{d}\sin(\theta)\\ \end{align} $$ and use Partial Fractions.