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If $\int \frac{(\cos x-\sin x)}{8-\sin 2 x} d x=\frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$, then $\mathrm{p}=($ Where $\mathrm{c}$ is a constant of integration)
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2818 Upvotes
Verified Answer
The correct answer is:
6
$$
\begin{aligned}
& \text { Let } I=\int \frac{\cos x-\sin x}{8-\sin 2 x} d x \\
& =\int \frac{\cos x-\sin x}{9-1-\sin 2 x} d x=\int \frac{\cos x-\sin x}{9-(1+\sin 2 x)} d x \\
& =\int \frac{\cos x-\sin x}{(3)^2-(\sin x+\cos x)^2} d x
\end{aligned}
$$
Put $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$
$$
\begin{aligned}
& \therefore I=\int \frac{d t}{(3)^2-(t)^2} d x=\frac{1}{3(2)} \log \left|\frac{3+t}{3-t}\right|+c \\
& =\frac{1}{6} \log \left|\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right|+c \Rightarrow p=6
\end{aligned}
$$
\begin{aligned}
& \text { Let } I=\int \frac{\cos x-\sin x}{8-\sin 2 x} d x \\
& =\int \frac{\cos x-\sin x}{9-1-\sin 2 x} d x=\int \frac{\cos x-\sin x}{9-(1+\sin 2 x)} d x \\
& =\int \frac{\cos x-\sin x}{(3)^2-(\sin x+\cos x)^2} d x
\end{aligned}
$$
Put $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$
$$
\begin{aligned}
& \therefore I=\int \frac{d t}{(3)^2-(t)^2} d x=\frac{1}{3(2)} \log \left|\frac{3+t}{3-t}\right|+c \\
& =\frac{1}{6} \log \left|\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right|+c \Rightarrow p=6
\end{aligned}
$$
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