# Indefinite integral $\displaystyle \int 1 / (p + q \sin a x)^2 \, \mathrm d x$

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.

$$q(1-u^2)=-\frac q p (pu^2+2qu+p) +\frac{2q}p( qu +p)$$
• @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 – Quanto Dec 30 '20 at 17:25