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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:

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

and:

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

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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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