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$\int \frac{\cos x-\sin x}{5+\sin (2 x)} d x=$
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1868 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2} \tan ^{-1}\left[\frac{1}{2}(\sin x+\cos x)\right]+c$
$$
\text { } \begin{aligned}
I & =\int \frac{\cos x-\sin x}{5+\sin 2 x} d x \\
& =\int \frac{\cos x-\sin x}{4+(\sin x+\cos x)^2} d x
\end{aligned}
$$
Let $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$
$$
\begin{aligned}
\therefore \quad I & =\int \frac{d t}{4+t^2}=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C \\
& =\frac{1}{2} \tan ^{-1}\left(\frac{\sin x+\cos x}{2}\right)+C
\end{aligned}
$$
\text { } \begin{aligned}
I & =\int \frac{\cos x-\sin x}{5+\sin 2 x} d x \\
& =\int \frac{\cos x-\sin x}{4+(\sin x+\cos x)^2} d x
\end{aligned}
$$
Let $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$
$$
\begin{aligned}
\therefore \quad I & =\int \frac{d t}{4+t^2}=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C \\
& =\frac{1}{2} \tan ^{-1}\left(\frac{\sin x+\cos x}{2}\right)+C
\end{aligned}
$$
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