MHT CET · Maths · Indefinite Integration
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)
- A 12
- B \(\frac{1}{6}\)
- C 6
- D 3
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { Let } I=\int \frac{\cos x-\sin x}{8-\sin 2 x} d x \\
& =\int \frac{\cos x-\sin x}{9-1-\sin 2 x} d x=\int \frac{\cos x-\sin x}{9-(1+\sin 2 x)} d x \\
& =\int \frac{\cos x-\sin x}{(3)^2-(\sin x+\cos x)^2} d x
\end{aligned}
\)
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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