Question

I= $\int \frac{(3\sin \theta -2)\cos \theta }{5-{\cos }^2\theta -4\sin \theta }d\theta$

Answer 1

$I=\int \frac{\left(3\sin \theta -2\right)}{5-{\cos }^2\theta -4\sin \theta }d\theta$

$=\int \frac{\left(3\sin \theta -2\right)\cos \theta }{5-\left(1-{\sin }^2\theta \right)-4\sin \theta }d\theta$

$=\int \frac{\left(3\sin \theta -2\right)\cos \theta }{5-1+{\sin }^2\theta -4\sin \theta }d\theta$

$=\int \frac{\left(3\sin \theta -2\right)\cos \theta }{{\sin }^2\theta -4\sin \theta +4}d\theta$

Putting $\sin \theta =t$

$\cos \theta d\theta =dt$

$I=\int \frac{\left(3t-2\right)}{t^2-4t+4}$

Putting $3t-2=A\frac{d}{dt}(t^2-4t+4)+B$

$3t-2=A(2t-4)+B$

$3t-2=2At-4A+B$

On comparing both side

$2A=3$ $-4A+B=-2$

$A=\frac{3}{2}$ $\Rightarrow$ $-4\frac{3}{2}+B=-2$

$\Rightarrow$ $-6+B=-2$

$B=4$

$I=\int \frac{\frac{3}{2}\left(2t-4\right)}{t^2-4t+4}dt+4\int \frac{dt}{t^2-4t+4}$

Putting $t^2-4t+4=y$

$(2t-4)dt=dy$

$\frac{3}{2}\int \frac{dt}{y}+4\int \frac{dt}{t^2-4t+4}$

$\frac{3}{2}\int \frac{dy}{y}+4\int \frac{dt}{{\left(t-2\right)}^2}$

$\frac{3}{2}\log \left|y\right|+4\frac{{\left(t-2\right)}^{-1}}{-1}+c$

$\frac{3}{2}\log \left|t^2-4t+4\right|-\frac{4}{t-2}+c$