Question

Show that $\int \frac{dx}{\sin x(5+4\cos x)}=\frac{1}{18}\log (1-\cos x)-\frac{1}{2}\log (1+\cos x)+\frac{4}{9}\log (5+4\cos x)$.

Answer 1

Let $I=\int \frac{1}{\sin x(5+4\cos x)}dx$
Multiplying numerator and denominator by $\sin x$, we get
$I=-\int \frac{\sin x}{(4\cos x+5)({\cos }^{2}x-1)}dx$
Put $t=\cos x$

$\Rightarrow dt=-\sin xdx$

So, $I=\int \frac{1}{(4t+5)(t^2-1)}dt=\frac{1}{(4t+5)(t+1)(t-1)}$

Resolving $\frac{1}{(4t+5)(t+1)(t-1)}$ into partial fractions

$\frac{1}{(4t+5)(t+1)(t-1)}=\frac{A}{4t+5}+\frac{B}{t+1}+\frac{C}{t-1}$ ....(1)

$\Rightarrow 1=A(t^2-1)+B(t-1)(4t+5)+C(4t+5)(t+1)$ ....(2)

Put $t=1$ in eqn (2)

$\Rightarrow C=\frac{1}{18}$

Put $t=-1$ ineqn (2)

$\Rightarrow B=-\frac{1}{2}$

Put $t=-\frac{5}{4}$

$\Rightarrow A=\frac{16}{9}$

Put these values in eqn (1)

$\frac{1}{(4t+5)(t+1)(t-1)}=\frac{16}{9(4t+5)}+\frac{-1}{2(t+1)}+\frac{1}{18(t-1)}$

$I=\int \frac{1}{(4t+5)(t^2-1)}dt=\frac{16}{9}\int \frac{dt}{(4t+5)}-\frac{1}{2}\int \frac{dt}{(t+1)}+\frac{1}{18}\int \frac{dt}{(t-1)}$
$=-\frac{1}{2}\log (t+1)+\frac{16}{9}\times \frac{1}{4}\log (4t+5)+\frac{1}{18}\log (t-1)+C$
$=-\frac{1}{2}\log (\cos x+1)+\frac{4}{9}\log (4\cos x+5)+\frac{1}{18}\log (\cos x-1)+C$