Question

If $\int \frac{(3\sin \varphi -2)\cos \varphi }{5-{\cos }^2\varphi -4\sin \varphi }d\varphi =3\ln (2-\sin \varphi )+\frac{k}{2+m\sin \varphi }+C$, then which of the following is/are true:
(where $C$ is integration constant)

• A. $k^2+1=16$
• B. $k^2-1=15$
• C. $m+1=0$
• D. $k+m=3$

Answer 1

Answer: (B), (C), (D)

$k+m=3$

$I=\int \frac{(3\sin \varphi -2)\cos \varphi }{5-{\cos }^2\varphi -4\sin \varphi }d\varphi$
Let $y=\sin \varphi \Rightarrow dy=\cos \varphi d\varphi$
Therefore
$\int \frac{(3\sin \varphi -2)\cos \varphi }{5-{\cos }^2\varphi -4\sin \varphi }d\varphi =\int \frac{(3y-2)dy}{5-(1-y^2)-4y}=\int \frac{(3y-2)}{y^2-4y+4}dy=\int \frac{(3y-2)}{(y-2{)}^2}$
Now, we can write
$\frac{3y-2}{(y-2{)}^2}=\frac{A}{y-2}+\frac{B}{(y-2{)}^2}$
Therefore
$3y-2=A(y-2)+B$
Comparing the coefficients of $y$ and constant term, we get
$A=3$ and $B=2A-2\Rightarrow B=4$
Therefore, the required integral is given by
$I=\int [\frac{3}{y-2}+\frac{4}{(y-2{)}^2}]dy=3\int \frac{dy}{y-2}+4\int \frac{dy}{(y-2{)}^2}=3\ln |y-2|-4\left(\frac{1}{y-2}\right)+C=3\ln |\sin \varphi -2|+\frac{4}{2-\sin \varphi }+C=3\ln (2-\sin \varphi )+\frac{4}{2-\sin \varphi }+C[\because (2-\sin \varphi \left)\text{is always positive}\right]$
$\therefore$ On comparing, we get
$k=4;m=-1$