Integrate $\int \frac {dx}{a+b\sin x}$ where $a^2<b^2$

Question

Integrate: $$\int \dfrac {dx}{a+b\sin x} \quad \,, \,a^2<b^2$$

My Attempt:

$$=\int \dfrac {dx}{a + b\dfrac {2\tan (\dfrac {x}{2})}{1+\tan^2 (\dfrac {x}{2})}}$$ $$=\int \dfrac {(1+\tan^2 (x/2)) dx}{a+a\tan^2 (x/2) + 2b\tan (x/2)}$$ $$=\dfrac {1}{a} \int \dfrac {\sec^2 (x/2) dx}{\tan^2 (x/2) + \dfrac {2b}{a} \tan (x/2) +1}$$ Put $\tan (\dfrac {x}{2})=t$ $$\sec^2 (x/2) dx=2 dt$$ Now, $$=\dfrac {2}{a} \int \dfrac {dt}{t^2+\dfrac {2b}{a} t +1}$$ $$=\dfrac {2}{a} \int \dfrac {dt}{(t+\dfrac {b}{a})^2 + \dfrac {a^2-b^2}{a^2}}$$

How to make use of the condition $a^2<b^2$ given in the question?

Answer 1

You're almost there:

Given that $a^2-b^2 < 0$,

$$\dfrac {2}{a} \int \dfrac {dt}{(t+\dfrac {b}{a})^2 + \dfrac {a^2-b^2}{a^2}} = \dfrac {2}{a} \int \dfrac {dt}{(t+\dfrac {b}{a})^2 -\alpha^2}$$

where $\alpha^2 = \dfrac {b^2-a^2}{a^2} > 0$.

Also let $t+\dfrac {b}{a} = v$, just to simplify things a bit. So therefore you're left with:

$$\dfrac {2}{a} \int \dfrac {dv}{v^2 - \alpha^2}$$

Now split the fraction:

$$\dfrac {2}{a} \int \dfrac {dv}{(v - \alpha)(v+\alpha)}$$

=$$\dfrac{2}{a \times 2 \alpha} \int \dfrac {(v+\alpha)-(v-\alpha)}{(v - \alpha)(v+\alpha)}dv$$

You can take it from here, yeah?

Answer 2

If your reasoning is correct, then you can factor the final integrand as $\frac{1}{((t+b/a)+\sqrt{b^2-a^2}/a)((t+b/a)-\sqrt{b^2-a^2}/a)}$

Answer 3

Well if $a^2\lt b^2$ then we have $$\frac{a^2-b^2}{a^2}\lt0$$ hence the integral is of the form $$\int\frac{\mathrm{d}x}{x^2-c^2}=-\frac1c\text{artanh}\left(\frac{x}c\right)$$

Answer 4

As, $a^2 < b^2$, $\frac{a^2 - b^2}{a^2} = -\frac{b^2 - a^2}{a^2} $

The final part has the form

$\int\frac{1}{(t+C)^2 - D^2}dt$

which yields, $\frac{1}{2D}ln\vert\frac{t+C-D}{t+C+D}\vert$

here, $D = \frac{\sqrt{b^2-a^2}}{|a|}, $ which requires the stated condition.