I am evaluating:

$$\int \frac {\mathrm d x} {(p + q \sin a x)^2}$$

which the book gives me as: $$\frac {q \cos a x} {a (p^2 - q^2) (p + q \sin a x) } + \frac p {p^2 - q^2} \int \frac {\mathrm d x} {p + q \sin a x}$$

This is Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.361$.

I am using a Weierstrass substitution: $u = \tan \dfrac {a x} 2$ which yields $\mathrm d x = \dfrac 2 a \dfrac {\mathrm d u} {1 + u^2}$ and $\sin a x = \dfrac {2 u} {u^2 + 1}$.

After algebra, this yields me the primitive:

$$\frac 2 a \int \frac {(u^2 + 1) \mathrm d u} {(p u^2 + 2 q u + p)^2}$$

I split this up into two separate primitives:

$$\frac 2 a \int \frac {u^2 \mathrm d u} {(p u^2 + 2 q u + p)^2} + \frac 2 a \int \frac {\mathrm d u} {(p u^2 + 2 q u + p)^2}$$

both of which are obtained via standard (though unwieldy) results:

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


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

After some straightforward cleaning up, I get:

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

The term on the right is in completely the correct format, returning me $\displaystyle \frac p {p^2 - q^2} \int \frac {\mathrm d x} {p + q \sin a x}$ after I put $u$ back.

But my left hand term has gone astray. In order to return $\dfrac {q \cos a x} {a (p^2 - q^2) (p + q \sin a x) }$, I really need to get it into the form:

$$\frac {q (1 - u^2)} {a (p^2 - q^2) (p u^2 + 2 q u + p)}$$

but instead I have: $$\frac {2 q (q u + p) } {a p (p^2 - q^2) (p u^2 + 2 q u + p)}$$

I have gone through my working carefully several times, but I can't put my finger on where I have gone astray -- it could be at any of the above steps.


Write the numerator as

$$q(1-u^2)=-\frac q p (pu^2+2qu+p) +\frac{2q}p( qu +p) $$

where the first term cancels the denominator, becoming a non-essential constant, and second term produces what you have.

  • $\begingroup$ Thank you for that. I wonder whether there's a better way to approach this than using a Weierstrass substitution, but every other approach I tried (various substitutions, splitting for parts in a number of ways) failed to get me anywhere. Hence Weierstrass it was, with all the complicated post-integration manipulation that resulted. $\endgroup$ – Prime Mover Dec 30 '20 at 17:12
  • 2
    $\begingroup$ @PrimeMover - My preference is to recognize $$\left(\frac{\cos x}{p+q\sin x}\right)’= \frac{p^2-q^2}{q(p+q\sin x)^2}- \frac pq \frac1{p+q\sin x}$$ which I know from experience $\endgroup$ – Quanto Dec 30 '20 at 17:25
  • $\begingroup$ Now I've managed to work my way through that, I understand how it works -- but I can't help but think that coming up with this sort of result is a bit of a dark art. $\endgroup$ – Prime Mover Dec 31 '20 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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