# Thread: Primitive of 1 / (1 - \sin a x)^2

1. ## Primitive of 1 / (1 - \sin a x)^2

Here we are again -- let's get cracking on some of these hard integrals again!

Number 358 in my source work:

$\int \frac {\mathrm d x} {\left({1 - \sin a x}\right)^2}$

which has to be demonstrated to be equal to:

$\frac 1 {2a} \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 1 {6 a} \tan^3 \left({\frac \pi 4 + \frac {a x} 2}\right) + C$

I tried a Weierstrass substitution: $u = \tan \frac {a x} 2$ but ended up with the horrible:

$\frac 2 a \int \frac {\left({u^2 + 1}\right) \ \mathrm d u} {\left({u - 1}\right)^4}$

which doesn't look like it's going in the right direction.

I also tried integration by parts, setting both $u$ and $\mathrm d v$ to $\frac 1 {1 - \sin a x}$ but got no further than:

$\frac 1 {1 - \sin a x} \left({\frac 1 a \tan \left({\frac \pi 4 + \frac {a x} 2}\right)}\right) - \int \left({\frac 1 a \tan \left({\frac \pi 4 + \frac {a x} 2}\right)}\right) \left({\frac {a \cos a x} {\left({1 - \sin a x}\right)^2} }\right) \ \mathrm d x$

... which looks horrible and I haven't had the courage to take it further down that route.

I tried a few substitutions but they didn't look promising either.

Any hints as to how this may be attacked?

2. ## Re: Primitive of 1 / (1 - \sin a x)^2

Try multiplying top and bottom by $(1+\sin ax)^2$

3. ## Re: Primitive of 1 / (1 - \sin a x)^2

Originally Posted by Jester
Try multiplying top and bottom by $(1+\sin ax)^2$
Looks tractable. I have integrals with $\cos^4$ on the bottom and various quantities of sine on the top which should be easy enough. Working through them presently.

4. ## Re: Primitive of 1 / (1 - \sin a x)^2

Originally Posted by Jester
Try multiplying top and bottom by $(1+\sin ax)^2$
THis gets me to:
$\frac {-2 \sin a x} {3 a \cos^3 a x} + \frac 1 {3 a} \tan a x + \frac 2 {3 a \cos^3 a x}$
from which I haven't got a clue as to how to get to:
$\frac 1 {2a} \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 1 {6 a} \tan^3 \left({\frac \pi 4 + \frac {a x} 2}\right)$

Presumably there are half-angle substitutions to be used, but one supposes there's an easier way to get there without going through this admittedly perfectly legitimate solution.

However, the form of the solution suggests that there may be integrations of secants to be found, as:
$\int \sec a x = \tan \left({\frac \pi 4 + \frac {a x} 2}\right)$

but I haven't been able to find it.

The reason I am going through this particular exercise is specifically to validate the results in this source work. Some of them appear to be wrong, and I want to make sure I know precisely where the errors are before I can recommend it as a learning aid.

5. ## Re: Primitive of 1 / (1 - \sin a x)^2

Originally Posted by Matt Westwood
THis gets me to:
$\frac {-2 \sin a x} {3 a \cos^3 a x} + \frac 1 {3 a} \tan a x + \frac 2 {3 a \cos^3 a x}$
from which I haven't got a clue as to how to get to:
$\frac 1 {2a} \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 1 {6 a} \tan^3 \left({\frac \pi 4 + \frac {a x} 2}\right)$

Presumably there are half-angle substitutions to be used, but one supposes there's an easier way to get there without going through this admittedly perfectly legitimate solution.

However, the form of the solution suggests that there may be integrations of secants to be found, as:
$\int \sec a x = \tan \left({\frac \pi 4 + \frac {a x} 2}\right)$

but I haven't been able to find it.

The reason I am going through this particular exercise is specifically to validate the results in this source work. Some of them appear to be wrong, and I want to make sure I know precisely where the errors are before I can recommend it as a learning aid.

D'oh!!!

This is rubbish, of course:

$\int \sec a x = \tan \left({\frac \pi 4 + \frac {a x} 2}\right)$

the correct statement is:

$\int \sec a x = \ln \left\vert{\tan \left({\frac \pi 4 + \frac {a x} 2}\right)}\right\vert$

6. ## Re: Primitive of 1 / (1 - \sin a x)^2

Actually I think I may have got this:

$\dfrac 1 {1 - \sin x} = \dfrac 1 2 \sec^2 \left({\dfrac \pi 4 + \dfrac x 2}\right)$

and take it from there.

7. ## Re: Primitive of 1 / (1 - \sin a x)^2

Looking at the answer (and more specifically the argument) I would suggest trying the substitution

$ax = 2u - \dfrac{\pi}{2}$

Funny, after using a trig identity, the integral becomes fairly easy.

8. ## Re: Primitive of 1 / (1 - \sin a x)^2

Originally Posted by Jester
Looking at the answer (and more specifically the argument) I would suggest trying the substitution

$ax = 2u - \dfrac{\pi}{2}$

Funny, after using a trig identity, the integral becomes fairly easy.
That's definitely worth a go, as a second method of solution.

Many thanks.

Now there's a whole load of similar integrals which will come tumbling. Always good to get a new technique on these things.