-1
$\begingroup$

This question already has an answer here:

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 ?

$\endgroup$

marked as duplicate by Eric Towers, Lord Shark the Unknown, user228113, RRL, Tom-Tom Mar 11 '18 at 21:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you realize how many duplicate questions there are to this? $\endgroup$ – Paul Mar 11 '18 at 18:43
  • $\begingroup$ Nope ... please give me some links...the rules might be same whereas the forms might be unseen by me . It would be helpful... $\endgroup$ – A.H.M. Mar 11 '18 at 18:44
  • $\begingroup$ @Paul Entering "\int\sec^3\theta d\theta" in the search box turned up literally dozens of pages; all I checked had this integral. $\endgroup$ – user5713492 Mar 11 '18 at 18:52
1
$\begingroup$

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 $\cos\theta\,\mathrm d\theta=\mathrm dx$

$\endgroup$
0
$\begingroup$

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 $m=3$!

$\endgroup$
  • $\begingroup$ Hmmm...can you explain a little broadly , please ... $\endgroup$ – A.H.M. Mar 11 '18 at 18:47
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Wrong on the first equation. $\endgroup$ – user5713492 Mar 11 '18 at 18:54
  • $\begingroup$ The integrand on the left is equal to $\sec^3$ and the integrand on the right is equal to $\sec^4$. $\endgroup$ – George Coote Mar 11 '18 at 19:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.