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 constant, then the range of $Af\left(x\right)$ is

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

Answer 1

The correct option is D $[-1,1]$

Let

$I=\int \frac{\sin (\frac{\pi }{4}-x)dx}{2+\sin 2x}$

Let $\frac{\pi }{4}-x=z\Rightarrow -dx=dz$

So,

$I=\int \frac{-\sin z}{2+\cos 2z}dz=\int \frac{(-\sin z)dz}{2{\cos }^{2}z+1}=\frac{-1}{2}\int \frac{\sin zdz}{{\cos }^{2}z+\frac{1}{2}}$

Let $\cos z=t$

So, $-\sin zdz=dt$

So,

$I=\frac{1}{2}\int \frac{dt}{t^2+\frac{1}{2}}=\frac{1}{2}\times \frac{\sqrt{2}}{1}\times {\tan }^{-1}\left(\frac{t\sqrt{2}}{1}\right)+CI=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}t\right)+C=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\cos z\right)+CI=\frac{1}{\sqrt{2}}{\tan }^{-1}\left[\sqrt{2}\cos (\frac{\pi }{4}-x)\right]+C=A{\tan }^{-1}\left(f\right(x\left)\right)+B$

So, $Af\left(x\right)=\frac{1}{\sqrt{2}}\times \sqrt{2}\cos (\frac{\pi }{4}-x)=\cos (\frac{\pi }{4}-x)$

and Range of $\cos (\frac{\pi }{4}-x)$ is $[-1,1]$