Integration of $\int\frac{1}{\cos x(5-4\sin x)}\, dx$

Question

$$\int\dfrac{1}{\cos x(5-4\sin x)}\, dx$$

I have tried multiple methods like converting this into partial fractions or substitution but no positive!

Answer 1

Alternative hint: $$ \frac{5+4\sin x}{\cos x}-\frac{16\cos x}{5-4\sin x}=\frac{25-16\sin^2x-16\cos^2x}{\cos x\cdot (5-4\sin x)}=\frac{9}{\cos x\cdot (5-4\sin x)} $$

Answer 2

Let $t=\tan\frac{x}{2}$.

Thus, $dt=\frac{1}{2}(1+t^2)dx$ and $$\int\frac{1}{\cos{x}(5-4\sin{x})}dx=\int\frac{\frac{2}{1+t^2}}{\frac{1-t^2}{1+t^2}\left(5-\frac{8t}{1+t^2}\right)}dt=$$ $$=\int\frac{2(1+t^2)}{(1-t^2)(5t^2-8t+5)}dt...$$ $$\frac{2(1+t^2)}{(1-t^2)(5t^2-8t+5)}=\frac{8}{9}\frac{5t-4}{5t^2-8t+5}-\frac{1}{t-1}+\frac{1}{9(1+t)}.$$

Answer 3

Hint: One way is $\sin x=u$ then $$\int\dfrac{1}{(\cos x)(5-4\sin x)}\,dx=\int\dfrac{\cos x\, dx}{(1-\sin^2x)(5-4\sin x)}=\int\dfrac{du}{(1-u^2)(5-4u)}$$

Answer 4

Let, $$Y=\int\dfrac{ 1}{cos(x)(5-4sin(x)} =\int \dfrac{ 1}{\dfrac{1-tan^2(x/2)}{1+tan^2(x/2)} × (5-4\dfrac{2tan(x/2)}{1+tan^2(x/2))})}$$ Now,simplify it and then put $tan(x/2)=t$

On differentiate bhoth side with d(t) I.e. d(tan(x/2))/d(t)=d(t)/d(t) sec^2(x/2)/2 × d(x)/d(t)=1 sec^2(x/2)d(x)=2d(t)