Question

If $\int \frac{2-3\sin x}{{\cos }^{2}x}dx=a\sec x+b\tan x+C,$ then which of the following is/are true (where $a$ and $b$ are fixed constants and $C$ is constant of integration)

• A. $b-a=5$
• B. $a-b=5$
• C. ${\sin }^{-1}\left(\frac{a}{b}\right)=-\frac{\pi }{3}$
• D. $\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$

Answer 1

Answer: (A), (D)

$\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$

By separating the terms, we get
$=\int (\frac{2}{{\cos }^{2}x}-\frac{3\sin x}{{\cos }^{2}x})dx=\int 2{\sec }^{2}xdx-3\int \tan x\sec xdx=2\tan x-3\sec x+C$
$\therefore a=-3,b=2$
$\Rightarrow b-a=5$
${\sin }^{-1}\left(\frac{a}{b}\right)={\sin }^{-1}\left(\frac{-3}{2}\right)$ is not defined
$\therefore \frac{1}{a}+\frac{1}{b}=-\frac{1}{3}+\frac{1}{2}=\frac{1}{6}$