Question

If $\int \frac{\sec x-\tan x}{\sqrt{{\sin }^{2}x-\sin x}}dx=k\ln |f\left(x\right)+\sqrt{2}\sqrt{\tan x(\tan x-\sec x)}|+c$, where $c$ is arbitrary constant and $k$ is a fixed constant, then

• A. $k=\sqrt{2}$
• B. $k=\frac{1}{\sqrt{2}}$
• C. $f\left(x\right)=\tan x-\sec x$
• D. $f\left(x\right)=\sqrt{\tan x+\sec x}$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)=\tan x-\sec x$
C $k=\sqrt{2}$

$I=\int \frac{\sec x-\tan x}{\sqrt{{\sin }^{2}x-\sin x}}dx$
$=\int \frac{(\sec x-\tan x)\sec x}{\sqrt{{\tan }^{2}x-\tan x\sec x}}dx$

Let $\tan x-\sec x=t$
$\Rightarrow \sec x(\sec x-\tan x)dx=dt$

Also, $(\tan x-\sec x{)}^2=t^2$
$\Rightarrow 2{\tan }^{2}x-2\sec x\tan x+1=t^2$
$\Rightarrow {\tan }^{2}x-\sec x\tan x=\frac{1}{2}(t^2-1)$
$\Rightarrow I=\int \frac{dt}{\sqrt{\frac{1}{2}(t^2-1)}}=\sqrt{2}\int \frac{dt}{\sqrt{t^2-1}}$
$=\sqrt{2}\ln |t+\sqrt{t^2-1}|+c$

On comparing with the given expression, we get
$k=\sqrt{2}$ and $f\left(x\right)=\tan x-\sec x$