Question

If $\int \frac{\cos x - \sin x}{8 - \sin 2x} \, dx = \frac{1}{p} \log \left[ \frac{3 + \sin x + \cos x}{3 - \sin x - \cos x} \right] + c$, then $p = $

• A. 12
• B. 16
• C. 6
• D. 3

Hint:

Whenever the numerator of a trigonometric fraction is $(\cos x - \sin x)$, always look to express your denominator in terms of $(\sin x + \cos x)^2$. This specific substitution simplifies the expressions into standard inverse quadratic formats instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given an integration identity containing a parameter $p$ in the denominator coefficient of the logarithmic solution. We need to evaluate the indefinite integral to determine the value of $p$.

Step 2: Key Formula or Approach:
Notice that the numerator $(\cos x - \sin x)$ is the exact derivative of the trigonometric sum $(\sin x + \cos x)$. This suggests using the substitution method: $$\text{Put } t = \sin x + \cos x \implies dt = (\cos x - \sin x) \, dx$$ To transform the denominator, square the substitution term to link it with $\sin 2x$: $$t^2 = (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x \implies \sin 2x = t^2 - 1$$

Step 3: Detailed Explanation:
Substitute the expression for $\sin 2x$ into the denominator of our integral: $$\text{Denominator } = 8 - \sin 2x = 8 - (t^2 - 1) = 8 - t^2 + 1 = 9 - t^2 = 3^2 - t^2$$ Now, rewrite the entire integral using the variable $t$: $$I = \int \frac{dt}{3^2 - t^2}$$ Apply the standard special inverse quadratic integral formula $\int \frac{dt}{a^2 - t^2} = \frac{1}{2a} \log \left| \frac{a + t}{a - t} \right| + c$: $$I = \frac{1}{2(3)} \log \left| \frac{3 + t}{3 - t} \right| + c = \frac{1}{6} \log \left| \frac{3 + \sin x + \cos x}{3 - \sin x - \cos x} \right| + c$$ Comparing this directly with the given solution format $\frac{1}{p} \log \left[ \dots \right] + c$: $$\frac{1}{p} = \frac{1}{6} \implies p = 6$$ This matches option (C).

Step 4: Final Answer:
The value of the parameter $p$ is 6, which corresponds to option (C).