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The value of $\sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)}$ is
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Verified Answer
The correct answer is:
$x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$
$x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$
$$
\begin{aligned}
& \sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)}=\sqrt{2} \int \frac{\sin \left(x-\frac{\pi}{4}+\frac{\pi}{4}\right) d x}{\sin \left(x-\frac{\pi}{4}\right)} \\
& =\sqrt{2} \int\left(\cos \frac{\pi}{4}+\cot \left(x-\frac{\pi}{4}\right) \sin \frac{\pi}{4}\right) d x \\
& =\int d x+\int \cot \left(x-\frac{\pi}{4}\right) d x \\
& =x+\ln \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c .
\end{aligned}
$$
\begin{aligned}
& \sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)}=\sqrt{2} \int \frac{\sin \left(x-\frac{\pi}{4}+\frac{\pi}{4}\right) d x}{\sin \left(x-\frac{\pi}{4}\right)} \\
& =\sqrt{2} \int\left(\cos \frac{\pi}{4}+\cot \left(x-\frac{\pi}{4}\right) \sin \frac{\pi}{4}\right) d x \\
& =\int d x+\int \cot \left(x-\frac{\pi}{4}\right) d x \\
& =x+\ln \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c .
\end{aligned}
$$
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