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If $\int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x=-\frac{1}{\sqrt{2}} \tan ^{-1}(f(x))+C$, then $f(x)=$
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Verified Answer
The correct answer is:
$\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)$
$$
\begin{aligned}
& I=\int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x=-\frac{1}{\sqrt{2}} \tan ^{-1}(f(x))+C \\
& I=\frac{1}{\sqrt{2}} \int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^2} d x
\end{aligned}
$$
On putting $\sin x+\cos x=t$
$$
\begin{aligned}
& (\cos x-\sin x) d x=d t \\
& \begin{aligned}
\Rightarrow \quad I & =-\frac{1}{\sqrt{2}} \int \frac{d t}{1+t^2}=-\frac{1}{\sqrt{2}} \tan ^{-1} t+C \\
& =-\frac{1}{\sqrt{2}} \tan ^{-1}(\sin x+\cos x)+C \\
\therefore f(x) & =\sin x+\cos x=\sqrt{2}\left(\cos \left(x-\frac{\pi}{4}\right)\right)
\end{aligned}
\end{aligned}
$$
\begin{aligned}
& I=\int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x=-\frac{1}{\sqrt{2}} \tan ^{-1}(f(x))+C \\
& I=\frac{1}{\sqrt{2}} \int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^2} d x
\end{aligned}
$$
On putting $\sin x+\cos x=t$
$$
\begin{aligned}
& (\cos x-\sin x) d x=d t \\
& \begin{aligned}
\Rightarrow \quad I & =-\frac{1}{\sqrt{2}} \int \frac{d t}{1+t^2}=-\frac{1}{\sqrt{2}} \tan ^{-1} t+C \\
& =-\frac{1}{\sqrt{2}} \tan ^{-1}(\sin x+\cos x)+C \\
\therefore f(x) & =\sin x+\cos x=\sqrt{2}\left(\cos \left(x-\frac{\pi}{4}\right)\right)
\end{aligned}
\end{aligned}
$$
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