Question

If ${\int }^{}\frac{\sin (\frac{\pi }{4}-x)}{2+\sin 2x}dx=A{\tan }^{-1}\left(f\right(x\left)\right)+B$, where $A$ and $B$ are constants, then the range of $Af\left(x\right)$ is

• A. $\left[\frac{-1}{2},\frac{1}{2}\right]$
• B. $[-1,0]$
• C. $[-\sqrt{2},\sqrt{2}]$
• D. $[-1.1]$

Answer 1

The correct option is A $\left[\frac{-1}{2},\frac{1}{2}\right]$

We have

$\int \frac{\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x}{2+\sin 2x}dx$

$=\frac{1}{\sqrt{2}}\int \frac{\left(\cos x-\sin x\right)}{2+{\left(\sin x+\cos x\right)}^2}dx$

$Let\sin x+\cos x=t$

Now

$\frac{1}{\sqrt{2}}\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}$

$=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+c$

$=\frac{1}{2}{\tan }^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)+c$

$A=\frac{1}{2}f\left(x\right)=\frac{\sin x+\cos x}{\sqrt{2}}$

$Af\left(x\right)=\frac{1}{2\sqrt{2}}\left(\sin x+\cos x\right)$

$\in \left[\frac{-\sqrt{2}}{2\sqrt{2}},\frac{\sqrt{2}}{2\sqrt{2}}\right]$

$\in \left[\frac{-1}{2},\frac{1}{2}\right]$

Hence, the option $\left(A\right)$ is the correct answer.