Evaluating $\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$

Question

How do I integrate this expression:

$$\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$$.I got this in a book.I do not know how to evaluate integrals of this type.

Answer 1

One uses trigonometric substitution: $t = \tan\left(\frac{x}{2}\right)$. Then $$ \sin(x) = \frac{2t}{1+t^2} \quad \cos(x) = \frac{1-t^2}{1+t^2} \quad \mathrm{d} x = \frac{2}{1+t^2} \mathrm{d} t $$ Hence: $$\begin{eqnarray} \int \frac{ \ell \sin(x) + m \cos(x)}{(a \sin(x)+ b \cos(x))^2} \mathrm{d}x &=& \int \frac{ \ell \frac{2 t}{1-t^2} + m \frac{1-t^2}{1+t^2}}{\left(a \frac{2 t}{1+t^2}+ b \frac{1-t^2}{1+t^2}\right)^2} \frac{2}{1+t^2}\mathrm{d}t \\ &=& \int \frac{2 \ell t + m (1-t^2)}{\left(2 a t + b (1-t^2)\right)^2} 2 \mathrm{d}t \end{eqnarray}$$ The resulting rational function can be integrated using partial fraction decomposition of the integrand, for example.

Answer 2

There is a universal solution, based on the Weierstrass substitution $t=\tan(x/2)$. We end up integrating a pretty awful but rational function of $t$. This is handled usually with partial fractions.

We get $dx=\frac{2\,dt}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$.

The same substitution works for any rational function of $\sin x$ and $\cos x$.

Answer 3

You can try Weierstrass substitution: $t = \tan \frac x 2$, which will convert the integrand into a rational function of $t$.