Question

$\int \frac{(3\sin \varphi -2)\cos \varphi }{(5-{\cos }^2\varphi -4\sin \varphi )}d\varphi$ is

• A. $3\ln (2-\sin \varphi )\frac{4}{(\sin \varphi -2)}+c$
• B. $3\ln (\sin \varphi -2)+\frac{4}{2-\sin \varphi }+c$
• C. $\ln (2-\sin \varphi )+\frac{4}{2-\sin \varphi }+c$
• D. $3\ln (2-\sin \varphi )+\frac{4}{(2-\sin \varphi )}+c$

Answer 1

The correct option is D $3\ln (2-\sin \varphi )+\frac{4}{(2-\sin \varphi )}+c$

$Let\sin \varphi =t,\cos \varphi d\varphi =dt$
$I=\int \frac{(3t-2)dt}{5-(1-t^2)-4t}=\frac{3t-2}{(t-2{)}^2}$
$Let\frac{3t-2}{(t-2{)}^2}=\frac{A}{t-2}+\frac{B}{(t-2{)}^2}$
$A=3,B=4$
$I=\int \frac{3}{t-2}dt+\int \frac{4}{(t-2{)}^2}dtI=3\ln (2-\sin \varphi )+\frac{4}{2-\sin \varphi }+c$